Lógica matemática y fundamentos (2017-18) - Contribuciones del usuario [es]

Relación 7

2018-04-08T12:28:35Z

Carmarria:

<source lang = "isar">
chapter {* R7: Deducción natural en lógica de primer orden *}

theory R7
imports Main
begin

text {*
Demostrar o refutar los siguientes lemas usando sólo las reglas
básicas de deducción natural de la lógica proposicional, de los
cuantificadores y de la igualdad:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P
· excluded_middel:(¬P ∨ P)

· allI: ⟦∀x. P x; P x ⟹ R⟧ ⟹ R
· allE: (⋀x. P x) ⟹ ∀x. P x
· exI: P x ⟹ ∃x. P x
· exE: ⟦∃x. P x; ⋀x. P x ⟹ Q⟧ ⟹ Q

· refl: t = t
· subst: ⟦s = t; P s⟧ ⟹ P t
· trans: ⟦r = s; s = t⟧ ⟹ r = t
· sym: s = t ⟹ t = s
· not_sym: t ≠ s ⟹ s ≠ t
· ssubst: ⟦t = s; P s⟧ ⟹ P t
· box_equals: ⟦a = b; a = c; b = d⟧ ⟹ a: = d
· arg_cong: x = y ⟹ f x = f y
· fun_cong: f = g ⟹ f x = g x
· cong: ⟦f = g; x = y⟧ ⟹ f x = g y
*}

text {*
Se usarán las reglas notnotI y mt que demostramos a continuación.
*}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
∀x. P x ⟶ Q x ⊢ (∀x. P x) ⟶ (∀x. Q x)
------------------------------------------------------------------ *}

lemma ejercicio_1a: carmarria
assumes 1: "∀x. P x ⟶ Q x"
shows "(∀x. P x) ⟶ (∀x. Q x)"
proof -
{assume 2: "! x. P x"
{fix a
have 3: "P a ⟶ Q a" using 1 by (rule allE)
have 4: "P a" using 2 by (rule allE)
have 5: "Q a" using 3 4 by (rule mp)
}
hence 6: "! x. Q x" by (rule allI)
}
thus "(! x. P x) ⟶ (! x. Q x)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
∃x. ¬(P x) ⊢ ¬(∀x. P x)
------------------------------------------------------------------ *}

lemma ejercicio_2a: carmarria
assumes 1: "∃x. ¬(P x)"
shows "¬(∀x. P x)"
proof -
obtain a where 2: "¬(P a)" using 1 by (rule exE)
{assume 3: "! x. P x"
have 4: "P a" using 3 by (rule allE)
have 5: False using 2 4 by (rule notE)
}
thus "¬(! x. P x)" by (rule notI)
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
∀x. P x ⊢ ∀y. P y
------------------------------------------------------------------ *}

lemma ejercicio_3a: carmarria
assumes 1: "∀x. P x"
shows "∀y. P y"
proof -
{fix a
have 2: "P a" using 1 by (rule allE)
}
thus "! y. P y" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
∀x. P x ⟶ Q x ⊢ (∀x. ¬(Q x)) ⟶ (∀x. ¬ (P x))
------------------------------------------------------------------ *}

lemma ejercicio_4: carmarria
assumes 1: "∀x. P x ⟶ Q x"
shows "(∀x. ¬(Q x)) ⟶ (∀x. ¬ (P x))"
proof -
{assume 2: "!x. ¬(Q x)"
{fix a
have 3: "¬(Q a)" using 2 by (rule allE)
have 4: "P a ⟶ Q a" using 1 by (rule allE)
have 5: "¬(P a)" using 4 3 by (rule mt)
}
hence 6: "!x. ¬ (P x)" by (rule allI)
}
thus "(!x. ¬(Q x)) ⟶ (!x. ¬ (P x))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
∀x. P x ⟶ ¬(Q x) ⊢ ¬(∃x. P x ∧ Q x)
------------------------------------------------------------------ *}

lemma ejercicio_5: carmarria
assumes 1: "∀x. P x ⟶ ¬(Q x)"
shows "¬(∃x. P x ∧ Q x)"
proof -
{assume 2: "? x. P x & Q x"
obtain a where 3: "P a & Q a" using 2 by (rule exE)
have 4: "P a ⟶ ¬(Q a)" using 1 by (rule allE)
have 5: "P a" using 3 by (rule conjunct1)
have 6: "Q a" using 3 by (rule conjunct2)
have 7: "¬(Q a)" using 4 5 by (rule mp)
have 8: False using 7 6 by (rule notE)
}
thus "¬(? x. P x & Q x)" by (rule notI)
qed

text {* ---------------------------------------------------------------
Ejercicio 6. Demostrar
∀x y. P x y ⊢ ∀u v. P u v
------------------------------------------------------------------ *}

lemma ejercicio_6: carmarria
assumes 1: "∀x y. P x y"
shows "∀u v. P u v"
proof -
{fix a
have 2: "! y. P a y" using 1 by (rule allE)
{fix b
have 3: "P a b" using 2 by (rule allE)
}
hence 4: "!v. P a v" by (rule allI)
}
thus "!u v. P u v" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
∃x y. P x y ⟹ ∃u v. P u v
------------------------------------------------------------------ *}

lemma ejercicio_7: carmarria
assumes 1: "∃x y. P x y"
shows "∃u v. P u v"
proof -
obtain a where 2: "? y. P a y" using 1 by (rule exE)
obtain b where 3: "P a b" using 2 by (rule exE)
have 4: "? v. P a v" using 3 by (rule exI)
show 5: "? u v. P u v" using 4 by (rule exI)
qed

text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
∃x. ∀y. P x y ⊢ ∀y. ∃x. P x y
------------------------------------------------------------------ *}

lemma ejercicio_8: carmarria
assumes 1: "∃x. ∀y. P x y"
shows "∀y. ∃x. P x y"
proof -
obtain a where 2: "! y. P a y" using 1 by (rule exE)
{fix b
have 3: "P a b" using 2 by (rule allE)
have 4: "? x. P x b" using 3 by (rule exI)
}
thus "! y. ? x. P x y" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
∃x. P a ⟶ Q x ⊢ P a ⟶ (∃x. Q x)
------------------------------------------------------------------ *}

lemma ejercicio_9: carmarria
assumes 1: "∃x. P a ⟶ Q x"
shows "P a ⟶ (∃x. Q x)"
proof -
{assume 2: "P a"
obtain b where 3: "P a ⟶ Q b" using 1 by (rule exE)
have 4: "Q b" using 3 2 by (rule mp)
have 5: "? x. Q x" using 4 by (rule exI)
}
thus "P a ⟶ (? x. Q x)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar
P a ⟶ (∃x. Q x) ⊢ ∃x. P a ⟶ Q x
------------------------------------------------------------------ *}

lemma ejercicio_10a: carmarria
fixes P Q :: "'b ⇒ bool"
assumes 1: "P a ⟶ (∃x. Q x)"
shows "∃x. P a ⟶ Q x"
proof -
have 2: "¬(P a) ∨ (P a)" by (rule excluded_middle)
moreover {assume 3: "¬(P a)"
{assume 4: "P a"
have 5: False using 3 4 by (rule notE)
have 6: "Q b" using 5 by (rule FalseE)
}
hence 7: "P a ⟶Q b" by (rule impI)
have 8: "? x. P a ⟶ Q x" using 7 by (rule exI)
}
moreover {assume 9: "P a"
have 10: "? x. Q x" using 1 9 by (rule mp)
obtain b where 11: "Q b" using 10 by (rule exE)
{assume 12: "P a"
have 13: "Q b" using 11 by this
}
hence 13: "P a ⟶ Q b" by (rule impI)
have 14: "? x. P a ⟶ Q x" using 13 by (rule exI)
}
ultimately show "? x. P a ⟶ Q x" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
(∃x. P x) ⟶ Q a ⊢ ∀x. P x ⟶ Q a
------------------------------------------------------------------ *}

lemma ejercicio_11a: carmarria
assumes 1:"(∃x. P x) ⟶ Q a"
shows "∀x. P x ⟶ Q a"
proof-
{fix b
{assume 2: "P b"
have 3: "? x. P x" using 2 by (rule exI)
have 4: "Q a" using 1 3 by (rule mp)
}
hence 5: "P b ⟶ Q a" by (rule impI)
}
thus "! x. P x ⟶ Q a" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
∀x. P x ⟶ Q a ⊢ ∃ x. P x ⟶ Q a
------------------------------------------------------------------ *}

lemma ejercicio_12a: carmarria
assumes 1: "∀x. P x ⟶ Q a"
shows "∃x. P x ⟶ Q a"
proof -
have 2: "P b ⟶ Q a" using 1 by (rule allE)
show 3: "? x. P x ⟶ Q a" using 2 by (rule exI)
qed

text {* ---------------------------------------------------------------
Ejercicio 13. Demostrar
(∀x. P x) ∨ (∀x. Q x) ⊢ ∀x. P x ∨ Q x
------------------------------------------------------------------ *}

lemma ejercicio_13a: carmarria
assumes 1: "(∀x. P x) ∨ (∀x. Q x)"
shows "∀x. P x ∨ Q x"
proof -
{fix a
have 2: "(! x. P x) ∨ (! x. Q x)" using 1 by this
moreover {assume 3:"! x. P x"
have 4: "P a" using 3 by (rule allE)
have 5: "P a | Q a" using 4 by (rule disjI1)
}
moreover {assume 6: "! x. Q x"
have 7: "Q a" using 6 by (rule allE)
have 8: "P a | Q a" using 7 by (rule disjI2)
}
ultimately have 9: "P a | Q a" by (rule disjE)
}
thus "! x. P x | Q x" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
∃x. P x ∧ Q x ⊢ (∃x. P x) ∧ (∃x. Q x)
------------------------------------------------------------------ *}

lemma ejercicio_14a: carmarria
assumes 1: "∃x. P x ∧ Q x"
shows "(∃x. P x) ∧ (∃x. Q x)"
proof -
obtain a where 2: "P a & Q a" using 1 by (rule exE)
have 3: "P a" using 2 by (rule conjunct1)
have 4: "Q a" using 2 by (rule conjunct2)
have 5: "? x. P x" using 3 by (rule exI)
have 6: "? x. Q x" using 4 by (rule exI)
show "(? x. P x) ∧ (? x. Q x)" using 5 6 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
∀x y. P y ⟶ Q x ⊢ (∃y. P y) ⟶ (∀x. Q x)
------------------------------------------------------------------ *}

lemma ejercicio_15a: carmarria
assumes 1: "∀x y. P y ⟶ Q x"
shows "(∃y. P y) ⟶ (∀x. Q x)"
proof -
{assume 2: "? y. P y"
{fix a
have 3: "! y. P y ⟶ Q a" using 1 by (rule allE)
obtain b where 4: "P b" using 2 by (rule exE)
have 5: "P b ⟶ Q a" using 3 by (rule allE)
have 6: "Q a" using 5 4 by (rule mp)
}
hence 7: "! x. Q x" by (rule allI)
}
thus "(? y. P y) ⟶ (! x. Q x)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
¬(∀x. ¬(P x)) ⊢ ∃x. P x
------------------------------------------------------------------ *}

lemma ejercicio_16a: carmarria
assumes 1: "¬(∀x. ¬(P x))"
shows "∃x. P x"
proof -
{assume 2: "¬(? x. P x)"
{fix b
{assume 3: "P b"
have 4: "? x. P x" using 3 by (rule exI)
have 5: False using 2 4 by (rule notE)
}
hence 6: "¬(P b)" by (rule notI)
}
hence 7: "! x. ¬(P x)" by (rule allI)
have 8: False using 1 7 by (rule notE)
}
hence 9: "¬¬(? x. P x)" by (rule notI)
show "? x. P x" using 9 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 17. Demostrar
∀x. ¬(P x) ⊢ ¬(∃x. P x)
------------------------------------------------------------------ *}

lemma ejercicio_17a: carmarria
assumes 1: "∀x. ¬(P x)"
shows "¬(∃x. P x)"
proof -
{assume 2: "? x. P x"
obtain a where 3: "P a" using 2 by (rule exE)
have 4: "¬(P a)" using 1 by (rule allE)
have 5: False using 4 3 by (rule notE)
}
thus "¬(? x. P x)" by (rule notI)
qed

text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
∃x. P x ⊢ ¬(∀x. ¬(P x))
------------------------------------------------------------------ *}

lemma ejercicio_18a: carmarria
assumes 1: "∃x. P x"
shows "¬(∀x. ¬(P x))"
proof-
{assume 2: "! x. ¬(P x)"
obtain a where 3: "P a" using 1 by (rule exE)
have 4: "¬(P a)" using 2 by (rule allE)
have 5: False using 4 3 by (rule notE)
}
thus "¬(! x. ¬(P x))" by (rule notI)
qed

text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
P a ⟶ (∀x. Q x) ⊢ ∀x. P a ⟶ Q x
------------------------------------------------------------------ *}

lemma ejercicio_19a: carmarria
assumes 1: "P a ⟶ (∀x. Q x)"
shows "∀x. P a ⟶ Q x"
proof -
{fix b
{assume 2: "P a"
have 3: "! x. Q x" using 1 2 by (rule mp)
have 4: "Q b" using 3 by (rule allE)
}
hence 5: "P a ⟶ Q b" by (rule impI)
}
thus "! x. P a ⟶ Q x" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
{∀x y z. R x y ∧ R y z ⟶ R x z,
∀x. ¬(R x x)}
⊢ ∀x y. R x y ⟶ ¬(R y x)
------------------------------------------------------------------ *}

lemma ejercicio_20a: carmarria
assumes 1: "∀x y z. R x y ∧ R y z ⟶ R x z" and
2: "∀x. ¬(R x x)"
shows "∀x y. R x y ⟶ ¬(R y x)"
proof -
{fix a
{fix b
{assume 3: "R a b"
have 4: "! y z. R a y & R y z ⟶ R a z" using 1 by (rule allE)
have 5: "! z. R a b & R b z ⟶ R a z" using 4 by (rule allE)
have 6: "R a b & R b a ⟶ R a a" using 5 by (rule allE)
have 7: "¬(R a a)" using 2 by (rule allE)
have 8: "¬(R a b & R b a)" using 6 7 by (rule mt)
{assume 9: "R b a"
have 10: "R a b & R b a" using 3 9 by (rule conjI)
have 11: False using 8 10 by (rule notE)
}
hence 12: "¬(R b a)" by (rule notI)
}
hence 13: "R a b ⟶ ¬(R b a)" by (rule impI)
}
hence 14: "! y. R a y ⟶ ¬(R y a)" by (rule allI)
}
thus "! x y. R x y ⟶ ¬(R y x)" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
{∀x. P x ∨ Q x, ∃x. ¬(Q x), ∀x. R x ⟶ ¬(P x)} ⊢ ∃x. ¬(R x)
------------------------------------------------------------------ *}

lemma ejercicio_21a: carmarria
assumes 1: "∀x. P x ∨ Q x" and
2: "∃x. ¬(Q x)" and
3: "∀x. R x ⟶ ¬(P x)"
shows "∃x. ¬(R x)"
proof -
obtain a where 4: "¬(Q a)" using 2 by (rule exE)
have 5: "P a | Q a" using 1 by (rule allE)
moreover {assume 6: "P a"
have 7: "¬¬(P a)" using 6 by (rule notnotI)
have 8: "R a ⟶ ¬(P a)" using 3 by (rule allE)
have 9: "¬(R a)" using 8 7 by (rule mt)
have 10: "? x. ¬(R x)" using 9 by (rule exI)
}
moreover {assume 10: "Q a"
have 11: False using 4 10 by (rule notE)
have 12: "? x. ¬(R x)" using 11 by (rule FalseE)
}
ultimately show "? x. ¬(R x)" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
{∀x. P x ⟶ Q x ∨ R x, ¬(∃x. P x ∧ R x)} ⊢ ∀x. P x ⟶ Q x
------------------------------------------------------------------ *}

lemma ejercicio_22a: carmarria
assumes 1: "∀x. P x ⟶ Q x ∨ R x" and
2: "¬(∃x. P x ∧ R x)"
shows "∀x. P x ⟶ Q x"
proof -
{fix a
{assume 3: "P a"
have 4: "P a ⟶ Q a | R a" using 1 by (rule allE)
have 5: "Q a | R a" using 4 3 by (rule mp)
moreover {assume 6: "Q a"
}
moreover {assume 7: "R a"
have 8: "P a & R a" using 3 7 by (rule conjI)
have 9: "? x. P x & R x" using 8 by (rule exI)
have 10: False using 2 9 by (rule notE)
have 11: "Q a" using 10 by (rule FalseE)
}
ultimately have 12: "Q a" by (rule disjE)
}
hence 13: "P a ⟶ Q a" by (rule impI)
}
thus "! x. P x ⟶ Q x" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
∃x y. R x y ∨ R y x ⊢ ∃x y. R x y
------------------------------------------------------------------ *}

lemma ejercicio_23a: carmarria
assumes 1: "∃x y. R x y ∨ R y x"
shows "∃x y. R x y"
proof -
obtain a where 2: "? y. R a y | R y a" using 1 by (rule exE)
obtain b where 3: "R a b | R b a" using 2 by (rule exE)
moreover {assume 4: "R a b"
have 5: "? y. R a y" using 4 by (rule exI)
have 6: "? x y. R x y" using 5 by (rule exI)
}
moreover {assume 7: "R b a"
have 8: "? y. R b y" using 7 by (rule exI)
have 9: "? x y. R x y" using 8 by (rule exI)
}
ultimately show "? x y. R x y" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)
------------------------------------------------------------------ *}

lemma ejercicio_24a: carmarria
"(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)"
proof -
{assume 1: "? x. ! y. P x y"
{fix b
obtain a where 2: "! y. P a y" using 1 by (rule exE)
have 3: "P a b" using 2 by (rule allE)
have 4: "? x. P x b" using 3 by (rule exI)
}
hence 5: "! y. ? x. P x y" by (rule allI)
}
thus "(? x. ! y. P x y) ⟶ (! y. ? x. P x y)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
(∀x. P x ⟶ Q) ⟷ ((∃x. P x) ⟶ Q)
------------------------------------------------------------------ *}

lemma ejercicio_25a: carmarria
"(∀x. P x ⟶ Q) ⟷ ((∃x. P x) ⟶ Q)"
proof (rule iffI)
{assume 1: "! x. P x ⟶ Q"
{assume 2: "? x. P x"
obtain a where 3: "P a" using 2 by (rule exE)
have 4: "P a ⟶ Q" using 1 by (rule allE)
have 5: Q using 4 3 by (rule mp)
}
thus 6: "(? x. P x) ⟶ Q" by (rule impI)
}
next
{assume 7: "(? x. P x) ⟶ Q"
{fix a
{assume 8: "P a"
have 9: "? x. P x" using 8 by (rule exI)
have 10: Q using 7 9 by (rule mp)
}
hence 11: "P a ⟶ Q" by (rule impI)
}
thus 12: "! x. P x ⟶ Q" by (rule allI)
}
qed

text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
((∀x. P x) ∧ (∀x. Q x)) ⟷ (∀x. P x ∧ Q x)
------------------------------------------------------------------ *}

lemma ejercicio_26a: carmarria
"((∀x. P x) ∧ (∀x. Q x)) ⟷ (∀x. P x ∧ Q x)"
proof (rule iffI)
{assume 1: "((! x. P x) ∧ (! x. Q x))"
have 2: "! x. P x" using 1 by (rule conjunct1)
have 3: "! x. Q x" using 1 by (rule conjunct2)
{fix a
have 4: "P a" using 2 by (rule allE)
have 5: "Q a" using 3 by (rule allE)
have 6: "P a & Q a" using 4 5 by (rule conjI)
}
thus 7: "! x. P x & Q x" by (rule allI)
}
next
{assume 8: "! x. P x & Q x"
{fix a
have 9: "P a & Q a" using 8 by (rule allE)
have 10: "P a" using 9 by (rule conjunct1)
}
hence 11: "! x. P x" by (rule allI)
{fix a
have 12: "P a & Q a" using 8 by (rule allE)
have 13: "Q a" using 12 by (rule conjunct2)
}
hence 14: "! x. Q x" by (rule allI)
show "(! x. P x) & (! x. Q x)" using 11 14 by (rule conjI)
}
qed

text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar o refutar
((∀x. P x) ∨ (∀x. Q x)) ⟷ (∀x. P x ∨ Q x)
------------------------------------------------------------------ *}

lemma ejercicio_27a: carmarria
"((∀x. P x) ∨ (∀x. Q x)) ⟷ (∀x. P x ∨ Q x)"
proof (rule iffI)
{assume 1: "(! x. P x) | (! x. Q x)"
moreover {assume 2: "! x. P x"
{fix a
have 3: "P a" using 2 by (rule allE)
have 4: "P a | Q a" using 3 by (rule disjI1)
}
hence 5: "! x. P x | Q x" by (rule allI)
}
moreover {assume 6: "! x. Q x"
{fix a
have 7: "Q a" using 6 by (rule allE)
have 8:"P a | Q a" using 7 by (rule disjI2)
}
hence 9: "! x. P x | Q x" by (rule allI)
}
ultimately show "! x. P x | Q x" by (rule disjE)
}
next
{assume 10: "! x. P x | Q x"
oops

(* Si tomamos el universo {1,2}, P(x): x=1, Q(x): x=2. Tenemos que es verdad que
(∀x. x=1 ∨ x=2), pero no se cumple ni (∀x. x=1) ni (∀x. x=2), por lo tanto el bicondicional es falso *)

text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar o refutar
((∃x. P x) ∨ (∃x. Q x)) ⟷ (∃x. P x ∨ Q x)
------------------------------------------------------------------ *}

lemma ejercicio_28a: carmarria
"((∃x. P x) ∨ (∃x. Q x)) ⟷ (∃x. P x ∨ Q x)"
proof (rule iffI)
{assume 1:"(? x. P x) | (? x. Q x)"
moreover {assume 2: "? x. P x"
obtain a where 3: "P a" using 2 by (rule exE)
have 4: "P a | Q a" using 3 by (rule disjI1)
have 5: "? x. P x | Q x" using 4 by (rule exI)
}
moreover {assume 6: "? x. Q x"
obtain a where 7: "Q a" using 6 by (rule exE)
have 8: "P a | Q a" using 7 by (rule disjI2)
have 9: "? x. P x | Q x" using 8 by (rule exI)
}
ultimately show "? x. P x | Q x" by (rule disjE)
}
next
{assume 10: "? x. P x | Q x"
obtain a where 11: "P a | Q a" using 10 by (rule exE)
moreover {assume 12: "P a"
have 13: "? x. P x" using 12 by (rule exI)
have 14: "(? x. P x) | (? x. Q x)" using 13 by (rule disjI1)
}
moreover {assume 15: "Q a"
have 16: "? x. Q x" using 15 by (rule exI)
have 17: "(? x. P x) | (? x. Q x)" using 16 by (rule disjI2)
}
ultimately show "(? x. P x) | (? x. Q x)" by (rule disjE)
}
qed

text {* ---------------------------------------------------------------
Ejercicio 29. Demostrar o refutar
(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)
------------------------------------------------------------------ *}

lemma ejercicio_29: carmarria
"(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)"
oops
(* Si tomamos el universo {1,2} y P(x,y): x=y, entonces tenemos que (∀x. ∃y. x=y) pero no se cumple
que (∃y. ∀x. x=y) *)

text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar o refutar
(¬(∀x. P x)) ⟷ (∃x. ¬P x)
------------------------------------------------------------------ *}

lemma ejercicio_30a: carmarria
"(¬(∀x. P x)) ⟷ (∃x. ¬P x)"
proof (rule iffI)
{assume 1: "¬(! x. P x)"
{assume 2: "¬(? x. ¬(P x))"
{fix a
{assume 3: "¬P a"
have 4: "? x. ¬P x" using 3 by (rule exI)
have 5: False using 2 4 by (rule notE)
}
hence 6: "¬(¬P a)" by (rule notI)
have 7: "P a" using 6 by (rule notnotD)
}
hence 8: "! x. P x" by (rule allI)
have 9: False using 1 8 by (rule notE)
}
hence 10: "¬¬(? x. ¬(P x))" by (rule notI)
show "? x. ¬(P x)" using 10 by (rule notnotD)
}
next
{assume 11: "? x. ¬P x"
obtain a where 12: "¬P a" using 11 by (rule exE)
{assume 13: "! x. P x"
have 14: "P a" using 13 by (rule allE)
have 15: False using 12 14 by (rule notE)
}
thus "¬(! x. P x)" by (rule notI)
}
qed

section {* Ejercicios sobre igualdad *}

text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar o refutar
P a ⟹ ∀x. x = a ⟶ P x
------------------------------------------------------------------ *}

lemma ejercicio_31a: carmarria
assumes 1: "P a"
shows "∀x. x = a ⟶ P x"
proof -
{fix b
{assume 2: "b = a"
have 3: "a = b" using 2 by (rule sym)
have 4: "P b" using 3 1 by (rule subst)
}
hence 5: "b = a ⟶ P b" by (rule impI)
}
thus "! x. x = a ⟶ P x" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar o refutar
∃x y. R x y ∨ R y x; ¬(∃x. R x x)⟧ ⟹ ∃x y. x ≠ y
------------------------------------------------------------------ *}

lemma ejercicio_32a: carmarria
fixes R :: "'c ⇒ 'c ⇒ bool"
assumes 1: "∃x y. R x y ∨ R y x" and
2: "¬(∃x. R x x)"
shows "∃(x::'c) y. x ≠ y"
proof -
obtain a where 3: "? y. R a y | R y a" using 1 by (rule exE)
obtain b where 4: "R a b | R b a" using 3 by (rule exE)
{assume 5: "a = b"
have 6: "R a b | R b a" using 4 by this
moreover {assume 7: "R a b"
have 8: "R b b" using 5 7 by (rule subst)
}
moreover {assume 9: "R b a"
have 10: "R b b" using 5 9 by (rule subst)
}
ultimately have 11: "R b b" by (rule disjE)
have 12: "? x. R x x" using 11 by (rule exI)
have 13: False using 2 12 by (rule notE)
}
hence 14: " a ≠ b " by (rule notI)
have 15: "? y. a ≠ y" using 14 by (rule exI)
show "? x y. (x ≠ y)" (* using 15 by (rule exI) *)
oops

text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar o refutar
{∀x. P a x x,
∀x y z. P x y z ⟶ P (f x) y (f z)}
⊢ P (f a) a (f a)
------------------------------------------------------------------ *}

lemma ejercicio_33a: carmarria
assumes 1: "∀x. P a x x" and
2: "∀x y z. P x y z ⟶ P (f x) y (f z)"
shows "P (f a) a (f a)"
proof -
have 3: "P a a a" using 1 by (rule allE)
have 4: "! y z. P a y z ⟶ P (f a) y (f z)" using 2 by (rule allE)
have 5: "! z. P a a z ⟶ P (f a) a (f z)" using 4 by (rule allE)
have 6: "P a a a ⟶ P (f a) a (f a)" using 5 by (rule allE)
show "P (f a) a (f a)" using 6 3 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar o refutar
{∀x. P a x x,
∀x y z. P x y z ⟶ P (f x) y (f z)⟧
⊢ ∃z. P (f a) z (f (f a))
------------------------------------------------------------------ *}

lemma ejercicio_34a: carmarria
assumes 1: "∀x. P a x x" and
2: "∀x y z. P x y z ⟶ P (f x) y (f z)"
shows "∃z. P (f a) z (f (f a))"
proof -
have 3: "P a (f a) (f a)" using 1 by (rule allE)
have 4: "! y z. P a y z ⟶ P (f a) y (f z)" using 2 by (rule allE)
have 5: "! z. P a (f a) z ⟶ P (f a) (f a) (f z)" using 4 by (rule allE)
have 6: "P a (f a) (f a) ⟶ P (f a) (f a) (f (f a))" using 5 by (rule allE)
have 7: "P (f a) (f a) (f (f a))" using 6 3 by (rule mp)
show 8: "? z. P (f a) z (f (f a))" using 7 by (rule exI)
qed

text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar o refutar
{∀y. Q a y,
∀x y. Q x y ⟶ Q (s x) (s y)}
⊢ ∃z. Qa z ∧ Q z (s (s a))
------------------------------------------------------------------ *}

lemma ejercicio_35a: carmarria
assumes 1: "∀y. Q a y" and
2: "∀x y. Q x y ⟶ Q (s x) (s y)"
shows "∃z. Q a z ∧ Q z (s (s a))"
proof-
have 3: "! y. Q a y ⟶ Q (s a) (s y)" using 2 by (rule allE)
have 4: "Q a (s a) ⟶ Q (s a) (s (s a))" using 3 by (rule allE)
have 5: "Q a (s a)" using 1 by (rule allE)
have 6: "Q (s a) (s (s a))" using 4 5 by (rule mp)
have 7: "Q a (s a) & Q (s a) (s (s a))" using 5 6 by (rule conjI)
show "? z. Q a z & Q z (s (s a))" using 7 by (rule exI)
qed

text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar o refutar
{x = f x, odd (f x)} ⊢ odd x
------------------------------------------------------------------ *}

lemma ejercicio_36a: carmarria
assumes 1: "x = f x" and
2: "odd (f x)"
shows "odd x"
proof -
have 3: "f x = x" using 1 by (rule sym)
show "odd x" using 3 2 by (rule subst)
qed

text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar o refutar
{x = f x, triple (f x) (f x) x} ⊢ triple x x x
------------------------------------------------------------------ *}

lemma ejercicio_37a: carmarria
assumes 1: "x = f x" and
2: "triple (f x) (f x) x"
shows "triple x x x"
proof -
have 3: "f x = x" using 1 by (rule sym)
show 4: "triple x x x" using 3 2 by (rule subst)
qed

end

</source>

Relación 7

2018-04-07T19:10:04Z

Carmarria:

<source lang = "isar">
chapter {* R7: Deducción natural en lógica de primer orden *}

theory R7
imports Main
begin

text {*
Demostrar o refutar los siguientes lemas usando sólo las reglas
básicas de deducción natural de la lógica proposicional, de los
cuantificadores y de la igualdad:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P
· excluded_middel:(¬P ∨ P)

· allI: ⟦∀x. P x; P x ⟹ R⟧ ⟹ R
· allE: (⋀x. P x) ⟹ ∀x. P x
· exI: P x ⟹ ∃x. P x
· exE: ⟦∃x. P x; ⋀x. P x ⟹ Q⟧ ⟹ Q

· refl: t = t
· subst: ⟦s = t; P s⟧ ⟹ P t
· trans: ⟦r = s; s = t⟧ ⟹ r = t
· sym: s = t ⟹ t = s
· not_sym: t ≠ s ⟹ s ≠ t
· ssubst: ⟦t = s; P s⟧ ⟹ P t
· box_equals: ⟦a = b; a = c; b = d⟧ ⟹ a: = d
· arg_cong: x = y ⟹ f x = f y
· fun_cong: f = g ⟹ f x = g x
· cong: ⟦f = g; x = y⟧ ⟹ f x = g y
*}

text {*
Se usarán las reglas notnotI y mt que demostramos a continuación.
*}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
∀x. P x ⟶ Q x ⊢ (∀x. P x) ⟶ (∀x. Q x)
------------------------------------------------------------------ *}

lemma ejercicio_1a: carmarria
assumes 1: "∀x. P x ⟶ Q x"
shows "(∀x. P x) ⟶ (∀x. Q x)"
proof -
{assume 2: "! x. P x"
{fix a
have 3: "P a ⟶ Q a" using 1 by (rule allE)
have 4: "P a" using 2 by (rule allE)
have 5: "Q a" using 3 4 by (rule mp)
}
hence 6: "! x. Q x" by (rule allI)
}
thus "(! x. P x) ⟶ (! x. Q x)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
∃x. ¬(P x) ⊢ ¬(∀x. P x)
------------------------------------------------------------------ *}

lemma ejercicio_2a: carmarria
assumes 1: "∃x. ¬(P x)"
shows "¬(∀x. P x)"
proof -
obtain a where 2: "¬(P a)" using 1 by (rule exE)
{assume 3: "! x. P x"
have 4: "P a" using 3 by (rule allE)
have 5: False using 2 4 by (rule notE)
}
thus "¬(! x. P x)" by (rule notI)
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
∀x. P x ⊢ ∀y. P y
------------------------------------------------------------------ *}

lemma ejercicio_3a: carmarria
assumes 1: "∀x. P x"
shows "∀y. P y"
proof -
{fix a
have 2: "P a" using 1 by (rule allE)
}
thus "! y. P y" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
∀x. P x ⟶ Q x ⊢ (∀x. ¬(Q x)) ⟶ (∀x. ¬ (P x))
------------------------------------------------------------------ *}

lemma ejercicio_4: carmarria
assumes 1: "∀x. P x ⟶ Q x"
shows "(∀x. ¬(Q x)) ⟶ (∀x. ¬ (P x))"
proof -
{assume 2: "!x. ¬(Q x)"
{fix a
have 3: "¬(Q a)" using 2 by (rule allE)
have 4: "P a ⟶ Q a" using 1 by (rule allE)
have 5: "¬(P a)" using 4 3 by (rule mt)
}
hence 6: "!x. ¬ (P x)" by (rule allI)
}
thus "(!x. ¬(Q x)) ⟶ (!x. ¬ (P x))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
∀x. P x ⟶ ¬(Q x) ⊢ ¬(∃x. P x ∧ Q x)
------------------------------------------------------------------ *}

lemma ejercicio_5: carmarria
assumes 1: "∀x. P x ⟶ ¬(Q x)"
shows "¬(∃x. P x ∧ Q x)"
proof -
{assume 2: "? x. P x & Q x"
obtain a where 3: "P a & Q a" using 2 by (rule exE)
have 4: "P a ⟶ ¬(Q a)" using 1 by (rule allE)
have 5: "P a" using 3 by (rule conjunct1)
have 6: "Q a" using 3 by (rule conjunct2)
have 7: "¬(Q a)" using 4 5 by (rule mp)
have 8: False using 7 6 by (rule notE)
}
thus "¬(? x. P x & Q x)" by (rule notI)
qed

text {* ---------------------------------------------------------------
Ejercicio 6. Demostrar
∀x y. P x y ⊢ ∀u v. P u v
------------------------------------------------------------------ *}

lemma ejercicio_6: carmarria
assumes 1: "∀x y. P x y"
shows "∀u v. P u v"
proof -
{fix a
have 2: "! y. P a y" using 1 by (rule allE)
{fix b
have 3: "P a b" using 2 by (rule allE)
}
hence 4: "!v. P a v" by (rule allI)
}
thus "!u v. P u v" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
∃x y. P x y ⟹ ∃u v. P u v
------------------------------------------------------------------ *}

lemma ejercicio_7: carmarria
assumes 1: "∃x y. P x y"
shows "∃u v. P u v"
proof -
obtain a where 2: "? y. P a y" using 1 by (rule exE)
obtain b where 3: "P a b" using 2 by (rule exE)
have 4: "? v. P a v" using 3 by (rule exI)
show 5: "? u v. P u v" using 4 by (rule exI)
qed

text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
∃x. ∀y. P x y ⊢ ∀y. ∃x. P x y
------------------------------------------------------------------ *}

lemma ejercicio_8: carmarria
assumes 1: "∃x. ∀y. P x y"
shows "∀y. ∃x. P x y"
proof -
obtain a where 2: "! y. P a y" using 1 by (rule exE)
{fix b
have 3: "P a b" using 2 by (rule allE)
have 4: "? x. P x b" using 3 by (rule exI)
}
thus "! y. ? x. P x y" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
∃x. P a ⟶ Q x ⊢ P a ⟶ (∃x. Q x)
------------------------------------------------------------------ *}

lemma ejercicio_9: carmarria
assumes 1: "∃x. P a ⟶ Q x"
shows "P a ⟶ (∃x. Q x)"
proof -
{assume 2: "P a"
obtain b where 3: "P a ⟶ Q b" using 1 by (rule exE)
have 4: "Q b" using 3 2 by (rule mp)
have 5: "? x. Q x" using 4 by (rule exI)
}
thus "P a ⟶ (? x. Q x)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar
P a ⟶ (∃x. Q x) ⊢ ∃x. P a ⟶ Q x
------------------------------------------------------------------ *}

lemma ejercicio_10a: carmarria
fixes P Q :: "'b ⇒ bool"
assumes 1: "P a ⟶ (∃x. Q x)"
shows "∃x. P a ⟶ Q x"
proof -
have 2: "¬(P a) ∨ (P a)" by (rule excluded_middle)
moreover {assume 3: "¬(P a)"
{assume 4: "P a"
have 5: False using 3 4 by (rule notE)
have 6: "Q b" using 5 by (rule FalseE)
}
hence 7: "P a ⟶Q b" by (rule impI)
have 8: "? x. P a ⟶ Q x" using 7 by (rule exI)
}
moreover {assume 9: "P a"
have 10: "? x. Q x" using 1 9 by (rule mp)
obtain b where 11: "Q b" using 10 by (rule exE)
{assume 12: "P a"
have 13: "Q b" using 11 by this
}
hence 13: "P a ⟶ Q b" by (rule impI)
have 14: "? x. P a ⟶ Q x" using 13 by (rule exI)
}
ultimately show "? x. P a ⟶ Q x" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
(∃x. P x) ⟶ Q a ⊢ ∀x. P x ⟶ Q a
------------------------------------------------------------------ *}

lemma ejercicio_11a: carmarria
assumes 1:"(∃x. P x) ⟶ Q a"
shows "∀x. P x ⟶ Q a"
proof-
{fix b
{assume 2: "P b"
have 3: "? x. P x" using 2 by (rule exI)
have 4: "Q a" using 1 3 by (rule mp)
}
hence 5: "P b ⟶ Q a" by (rule impI)
}
thus "! x. P x ⟶ Q a" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
∀x. P x ⟶ Q a ⊢ ∃ x. P x ⟶ Q a
------------------------------------------------------------------ *}

lemma ejercicio_12a: carmarria
assumes 1: "∀x. P x ⟶ Q a"
shows "∃x. P x ⟶ Q a"
proof -
have 2: "P b ⟶ Q a" using 1 by (rule allE)
show 3: "? x. P x ⟶ Q a" using 2 by (rule exI)
qed

text {* ---------------------------------------------------------------
Ejercicio 13. Demostrar
(∀x. P x) ∨ (∀x. Q x) ⊢ ∀x. P x ∨ Q x
------------------------------------------------------------------ *}

lemma ejercicio_13a: carmarria
assumes 1: "(∀x. P x) ∨ (∀x. Q x)"
shows "∀x. P x ∨ Q x"
proof -
{fix a
have 2: "(! x. P x) ∨ (! x. Q x)" using 1 by this
moreover {assume 3:"! x. P x"
have 4: "P a" using 3 by (rule allE)
have 5: "P a | Q a" using 4 by (rule disjI1)
}
moreover {assume 6: "! x. Q x"
have 7: "Q a" using 6 by (rule allE)
have 8: "P a | Q a" using 7 by (rule disjI2)
}
ultimately have 9: "P a | Q a" by (rule disjE)
}
thus "! x. P x | Q x" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
∃x. P x ∧ Q x ⊢ (∃x. P x) ∧ (∃x. Q x)
------------------------------------------------------------------ *}

lemma ejercicio_14a: carmarria
assumes 1: "∃x. P x ∧ Q x"
shows "(∃x. P x) ∧ (∃x. Q x)"
proof -
obtain a where 2: "P a & Q a" using 1 by (rule exE)
have 3: "P a" using 2 by (rule conjunct1)
have 4: "Q a" using 2 by (rule conjunct2)
have 5: "? x. P x" using 3 by (rule exI)
have 6: "? x. Q x" using 4 by (rule exI)
show "(? x. P x) ∧ (? x. Q x)" using 5 6 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
∀x y. P y ⟶ Q x ⊢ (∃y. P y) ⟶ (∀x. Q x)
------------------------------------------------------------------ *}

lemma ejercicio_15a: carmarria
assumes 1: "∀x y. P y ⟶ Q x"
shows "(∃y. P y) ⟶ (∀x. Q x)"
proof -
{assume 2: "? y. P y"
{fix a
have 3: "! y. P y ⟶ Q a" using 1 by (rule allE)
obtain b where 4: "P b" using 2 by (rule exE)
have 5: "P b ⟶ Q a" using 3 by (rule allE)
have 6: "Q a" using 5 4 by (rule mp)
}
hence 7: "! x. Q x" by (rule allI)
}
thus "(? y. P y) ⟶ (! x. Q x)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
¬(∀x. ¬(P x)) ⊢ ∃x. P x
------------------------------------------------------------------ *}

lemma ejercicio_16a: carmarria
assumes 1: "¬(∀x. ¬(P x))"
shows "∃x. P x"
proof -
{assume 2: "¬(? x. P x)"
{fix b
{assume 3: "P b"
have 4: "? x. P x" using 3 by (rule exI)
have 5: False using 2 4 by (rule notE)
}
hence 6: "¬(P b)" by (rule notI)
}
hence 7: "! x. ¬(P x)" by (rule allI)
have 8: False using 1 7 by (rule notE)
}
hence 9: "¬¬(? x. P x)" by (rule notI)
show "? x. P x" using 9 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 17. Demostrar
∀x. ¬(P x) ⊢ ¬(∃x. P x)
------------------------------------------------------------------ *}

lemma ejercicio_17a: carmarria
assumes 1: "∀x. ¬(P x)"
shows "¬(∃x. P x)"
proof -
{assume 2: "? x. P x"
obtain a where 3: "P a" using 2 by (rule exE)
have 4: "¬(P a)" using 1 by (rule allE)
have 5: False using 4 3 by (rule notE)
}
thus "¬(? x. P x)" by (rule notI)
qed

text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
∃x. P x ⊢ ¬(∀x. ¬(P x))
------------------------------------------------------------------ *}

lemma ejercicio_18a: carmarria
assumes 1: "∃x. P x"
shows "¬(∀x. ¬(P x))"
proof-
{assume 2: "! x. ¬(P x)"
obtain a where 3: "P a" using 1 by (rule exE)
have 4: "¬(P a)" using 2 by (rule allE)
have 5: False using 4 3 by (rule notE)
}
thus "¬(! x. ¬(P x))" by (rule notI)
qed

text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
P a ⟶ (∀x. Q x) ⊢ ∀x. P a ⟶ Q x
------------------------------------------------------------------ *}

lemma ejercicio_19a: carmarria
assumes 1: "P a ⟶ (∀x. Q x)"
shows "∀x. P a ⟶ Q x"
proof -
{fix b
{assume 2: "P a"
have 3: "! x. Q x" using 1 2 by (rule mp)
have 4: "Q b" using 3 by (rule allE)
}
hence 5: "P a ⟶ Q b" by (rule impI)
}
thus "! x. P a ⟶ Q x" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
{∀x y z. R x y ∧ R y z ⟶ R x z,
∀x. ¬(R x x)}
⊢ ∀x y. R x y ⟶ ¬(R y x)
------------------------------------------------------------------ *}

lemma ejercicio_20a: carmarria
assumes 1: "∀x y z. R x y ∧ R y z ⟶ R x z" and
2: "∀x. ¬(R x x)"
shows "∀x y. R x y ⟶ ¬(R y x)"
proof -
{fix a
{fix b
{assume 3: "R a b"
have 4: "! y z. R a y & R y z ⟶ R a z" using 1 by (rule allE)
have 5: "! z. R a b & R b z ⟶ R a z" using 4 by (rule allE)
have 6: "R a b & R b a ⟶ R a a" using 5 by (rule allE)
have 7: "¬(R a a)" using 2 by (rule allE)
have 8: "¬(R a b & R b a)" using 6 7 by (rule mt)
{assume 9: "R b a"
have 10: "R a b & R b a" using 3 9 by (rule conjI)
have 11: False using 8 10 by (rule notE)
}
hence 12: "¬(R b a)" by (rule notI)
}
hence 13: "R a b ⟶ ¬(R b a)" by (rule impI)
}
hence 14: "! y. R a y ⟶ ¬(R y a)" by (rule allI)
}
thus "! x y. R x y ⟶ ¬(R y x)" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
{∀x. P x ∨ Q x, ∃x. ¬(Q x), ∀x. R x ⟶ ¬(P x)} ⊢ ∃x. ¬(R x)
------------------------------------------------------------------ *}

lemma ejercicio_21a: carmarria
assumes 1: "∀x. P x ∨ Q x" and
2: "∃x. ¬(Q x)" and
3: "∀x. R x ⟶ ¬(P x)"
shows "∃x. ¬(R x)"
proof -
obtain a where 4: "¬(Q a)" using 2 by (rule exE)
have 5: "P a | Q a" using 1 by (rule allE)
moreover {assume 6: "P a"
have 7: "¬¬(P a)" using 6 by (rule notnotI)
have 8: "R a ⟶ ¬(P a)" using 3 by (rule allE)
have 9: "¬(R a)" using 8 7 by (rule mt)
have 10: "? x. ¬(R x)" using 9 by (rule exI)
}
moreover {assume 10: "Q a"
have 11: False using 4 10 by (rule notE)
have 12: "? x. ¬(R x)" using 11 by (rule FalseE)
}
ultimately show "? x. ¬(R x)" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
{∀x. P x ⟶ Q x ∨ R x, ¬(∃x. P x ∧ R x)} ⊢ ∀x. P x ⟶ Q x
------------------------------------------------------------------ *}

lemma ejercicio_22a: carmarria
assumes 1: "∀x. P x ⟶ Q x ∨ R x" and
2: "¬(∃x. P x ∧ R x)"
shows "∀x. P x ⟶ Q x"
proof -
{fix a
{assume 3: "P a"
have 4: "P a ⟶ Q a | R a" using 1 by (rule allE)
have 5: "Q a | R a" using 4 3 by (rule mp)
moreover {assume 6: "Q a"
}
moreover {assume 7: "R a"
have 8: "P a & R a" using 3 7 by (rule conjI)
have 9: "? x. P x & R x" using 8 by (rule exI)
have 10: False using 2 9 by (rule notE)
have 11: "Q a" using 10 by (rule FalseE)
}
ultimately have 12: "Q a" by (rule disjE)
}
hence 13: "P a ⟶ Q a" by (rule impI)
}
thus "! x. P x ⟶ Q x" by (rule allI)
qed

text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
∃x y. R x y ∨ R y x ⊢ ∃x y. R x y
------------------------------------------------------------------ *}

lemma ejercicio_23a: carmarria
assumes 1: "∃x y. R x y ∨ R y x"
shows "∃x y. R x y"
proof -
obtain a where 2: "? y. R a y | R y a" using 1 by (rule exE)
obtain b where 3: "R a b | R b a" using 2 by (rule exE)
moreover {assume 4: "R a b"
have 5: "? y. R a y" using 4 by (rule exI)
have 6: "? x y. R x y" using 5 by (rule exI)
}
moreover {assume 7: "R b a"
have 8: "? y. R b y" using 7 by (rule exI)
have 9: "? x y. R x y" using 8 by (rule exI)
}
ultimately show "? x y. R x y" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)
------------------------------------------------------------------ *}

lemma ejercicio_24a: carmarria
"(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)"
proof -
{assume 1: "? x. ! y. P x y"
{fix b
obtain a where 2: "! y. P a y" using 1 by (rule exE)
have 3: "P a b" using 2 by (rule allE)
have 4: "? x. P x b" using 3 by (rule exI)
}
hence 5: "! y. ? x. P x y" by (rule allI)
}
thus "(? x. ! y. P x y) ⟶ (! y. ? x. P x y)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
(∀x. P x ⟶ Q) ⟷ ((∃x. P x) ⟶ Q)
------------------------------------------------------------------ *}

lemma ejercicio_25a: carmarria
"(∀x. P x ⟶ Q) ⟷ ((∃x. P x) ⟶ Q)"
proof (rule iffI)
{assume 1: "! x. P x ⟶ Q"
{assume 2: "? x. P x"
obtain a where 3: "P a" using 2 by (rule exE)
have 4: "P a ⟶ Q" using 1 by (rule allE)
have 5: Q using 4 3 by (rule mp)
}
thus 6: "(? x. P x) ⟶ Q" by (rule impI)
}
next
{assume 7: "(? x. P x) ⟶ Q"
{fix a
{assume 8: "P a"
have 9: "? x. P x" using 8 by (rule exI)
have 10: Q using 7 9 by (rule mp)
}
hence 11: "P a ⟶ Q" by (rule impI)
}
thus 12: "! x. P x ⟶ Q" by (rule allI)
}
qed

text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
((∀x. P x) ∧ (∀x. Q x)) ⟷ (∀x. P x ∧ Q x)
------------------------------------------------------------------ *}

lemma ejercicio_26a: carmarria
"((∀x. P x) ∧ (∀x. Q x)) ⟷ (∀x. P x ∧ Q x)"
proof (rule iffI)
{assume 1: "((! x. P x) ∧ (! x. Q x))"
have 2: "! x. P x" using 1 by (rule conjunct1)
have 3: "! x. Q x" using 1 by (rule conjunct2)
{fix a
have 4: "P a" using 2 by (rule allE)
have 5: "Q a" using 3 by (rule allE)
have 6: "P a & Q a" using 4 5 by (rule conjI)
}
thus 7: "! x. P x & Q x" by (rule allI)
}
next
{assume 8: "! x. P x & Q x"
{fix a
have 9: "P a & Q a" using 8 by (rule allE)
have 10: "P a" using 9 by (rule conjunct1)
}
hence 11: "! x. P x" by (rule allI)
{fix a
have 12: "P a & Q a" using 8 by (rule allE)
have 13: "Q a" using 12 by (rule conjunct2)
}
hence 14: "! x. Q x" by (rule allI)
show "(! x. P x) & (! x. Q x)" using 11 14 by (rule conjI)
}
qed

text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar o refutar
((∀x. P x) ∨ (∀x. Q x)) ⟷ (∀x. P x ∨ Q x)
------------------------------------------------------------------ *}

lemma ejercicio_27a:
"((∀x. P x) ∨ (∀x. Q x)) ⟷ (∀x. P x ∨ Q x)"
oops

text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar o refutar
((∃x. P x) ∨ (∃x. Q x)) ⟷ (∃x. P x ∨ Q x)
------------------------------------------------------------------ *}

lemma ejercicio_28a:
"((∃x. P x) ∨ (∃x. Q x)) ⟷ (∃x. P x ∨ Q x)"
oops

text {* ---------------------------------------------------------------
Ejercicio 29. Demostrar o refutar
(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)
------------------------------------------------------------------ *}

lemma ejercicio_29:

"(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)"
oops

text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar o refutar
(¬(∀x. P x)) ⟷ (∃x. ¬P x)
------------------------------------------------------------------ *}

lemma ejercicio_30a:
"(¬(∀x. P x)) ⟷ (∃x. ¬P x)"
oops

section {* Ejercicios sobre igualdad *}

text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar o refutar
P a ⟹ ∀x. x = a ⟶ P x
------------------------------------------------------------------ *}

lemma ejercicio_31a:
assumes "P a"
shows "∀x. x = a ⟶ P x"
oops

text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar o refutar
∃x y. R x y ∨ R y x; ¬(∃x. R x x)⟧ ⟹ ∃x y. x ≠ y
------------------------------------------------------------------ *}

lemma ejercicio_32a:
fixes R :: "'c ⇒ 'c ⇒ bool"
assumes "∃x y. R x y ∨ R y x"
"¬(∃x. R x x)"
shows "∃(x::'c) y. x ≠ y"
oops

text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar o refutar
{∀x. P a x x,
∀x y z. P x y z ⟶ P (f x) y (f z)}
⊢ P (f a) a (f a)
------------------------------------------------------------------ *}

lemma ejercicio_33a:
assumes "∀x. P a x x"
"∀x y z. P x y z ⟶ P (f x) y (f z)"
shows "P (f a) a (f a)"
oops

text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar o refutar
{∀x. P a x x,
∀x y z. P x y z ⟶ P (f x) y (f z)⟧
⊢ ∃z. P (f a) z (f (f a))
------------------------------------------------------------------ *}

lemma ejercicio_34a:
assumes "∀x. P a x x"
"∀x y z. P x y z ⟶ P (f x) y (f z)"
shows "∃z. P (f a) z (f (f a))"
oops

text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar o refutar
{∀y. Q a y,
∀x y. Q x y ⟶ Q (s x) (s y)}
⊢ ∃z. Qa z ∧ Q z (s (s a))
------------------------------------------------------------------ *}

lemma ejercicio_35a:
assumes "∀y. Q a y"
"∀x y. Q x y ⟶ Q (s x) (s y)"
shows "∃z. Q a z ∧ Q z (s (s a))"
oops

text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar o refutar
{x = f x, odd (f x)} ⊢ odd x
------------------------------------------------------------------ *}

lemma ejercicio_36a:
assumes "x = f x" and
"odd (f x)"
shows "odd x"
oops

text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar o refutar
{x = f x, triple (f x) (f x) x} ⊢ triple x x x
------------------------------------------------------------------ *}

lemma ejercicio_37a:
assumes "x = f x" and
"triple (f x) (f x) x"
shows "triple x x x"
oops

end

</source>

Relación 7

2018-04-07T12:01:27Z

Carmarria:

Relación 7

2018-04-06T15:37:27Z

Carmarria:

Relación 4

2018-03-22T14:33:51Z

Carmarria:

<source lang = "isar">

chapter {* R4: Deducción natural proposicional *}

theory R4
imports Main
begin

text {*
---------------------------------------------------------------------
El objetivo de esta relación es demostrar cada uno de los ejercicios
usando sólo las reglas básicas de deducción natural de la lógica
proposicional (sin usar el método auto).

Las reglas básicas de la deducción natural son las siguientes:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· notnotI: P ⟹ ¬¬ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P
---------------------------------------------------------------------
*}

text {*
Se usarán las reglas notnotI y mt que demostramos a continuación. *}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

section {* Implicaciones *}

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
p ⟶ q, p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_1: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ q" and
2: "p"
shows "q"

proof -
show 1: "q" using 1 2 by (rule mp)
qed

lemma ej_1: joslopjim4
assumes "p --> q"
"p"
shows "q"
proof -
show "q" using assms(1) assms(2) by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
p ⟶ q, q ⟶ r, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_2: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ q" and
2: "q ⟶ r" and
3: "p"
shows "r"

proof-
have 4: "q" using 1 3 by (rule mp)
show "r" using 2 4 by (rule mp)
qed

lemma ej_2: joslopjim4
assumes "p ⟶ q"
"q ⟶ r"
"p"
shows "r"
proof -
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_3: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ (q ⟶ r)" and
2: "p ⟶ q" and
3: "p"
shows "r"

proof-
have 4: "q" using 2 3 by (rule mp)
have 5: "(q ⟶ r)" using 1 3 by (rule mp)
show "r" using 5 4 by (rule mp)

qed

lemma ej_3: joslopjim4
assumes "p ⟶ (q ⟶ r)"
"p ⟶ q"
"p"
shows "r"
proof -
have "q" using `p⟶q` `p` by (rule mp)
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
p ⟶ q, q ⟶ r ⊢ p ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_4: josrodjim2
assumes 1: "p ⟶ q" and
2: "q ⟶ r"
shows "p ⟶ r"

proof-
{ assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)}
hence 6: "p ⟶ r" using 3 5 by (rule impI)
show "p⟶r" using 6 by this

oops

lemma ejercicio_4: carmarria
assumes 1: "p ⟶ q" and
2: "q ⟶ r"
shows "p ⟶ r"
proof -
{ assume 3: p
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)
}
thus "p ⟶ r" by (rule impI)
qed
lemma ejercicio_4: inmbenber
assumes 1: "p ⟶ q" and
2: "q ⟶ r"
shows "p ⟶ r"
proof -
{ assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp) }
thus "p ⟶ r" by (rule impI)
qed

lemma ejercicio_4:marmedmar3
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof
assume "p"
with assms(1) have "q" by (rule mp)
with assms(2) show "r" by (rule mp)
qed

lemma ej_4: joslopjim4
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof (rule impI)
assume "p"
have "q" using assms(1) `p` by (rule mp)
show "r" using assms(2) `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_5: josrodjim2 (NO DA ERROR, PERO SI ALERTA)
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

proof -

{ assume 2: "q"
{assume 3: "p"
have 4: "q ⟶r" using 1 3 by (rule mp)
have 5: "r" using 4 2 by (rule mp)}
hence 6: "p⟶r" using 3 5 by (rule impI)}
hence 7: "q ⟶ (p ⟶ r)" using 2 6 by (rule impI)
show "q ⟶ (p ⟶ r)" using 7 by this

oops

lemma ejercicio_5: carmarria
assumes 1: "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

proof -
{assume 2: q
{assume 3: p
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "r" using 4 2 by (rule mp)
}
hence " p ⟶ r" by (rule impI)
}
thus "q ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_5: inmbenber
assumes 1: "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

proof -
{assume 2: "q"
{assume 3: "p"
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "r" using 4 2 by (rule mp) }
hence " p ⟶ r" by (rule impI) }
thus "q ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_5: marmedmar3
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"
proof
assume "q"
show "(p ⟶ r)"
proof
assume "p"
with assms(1) have "(q ⟶ r)" by (rule mp)
thus "r" using `q` by (rule mp)
qed
qed

lemma ej_5: joslopjim4
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"
proof (rule impI)
assume "q"
show "p⟶r"
proof (rule impI)
assume "p"
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 6. Demostrar
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_6:josrodjim2
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"

proof-

{assume 2: "p⟶q"
{assume 3: "p"
have 4: "q⟶r" using 1 3 by (rule mp)
have 5: "q" using 2 3 by (rule mp)
have 6: "r" using 4 5 by (rule mp)}
hence 7: "p⟶r" using 3 6 by (rule impI)}
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" using 2 7 by (rule impI)
show "(p ⟶ q) ⟶ (p ⟶ r)" using 8 by this
oops

lemma ejercicio_6: carmarria, inmbenber
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof -
{assume 2: "p ⟶ q"
{assume 3: "p"
have 4: "q" using 2 3 by (rule mp)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: "r" using 5 4 by (rule mp)
}
hence 7: "p ⟶ r" by (rule impI)
}
thus "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_6:marmedmar3
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof
assume 2: "(p ⟶ q)"
show "(p ⟶ r)"
proof
assume 3: "p"
have 4: "q" using 2 3 by (rule mp)
have 5: "(q ⟶ r)" using 1 3 by (rule mp)
show "r" using 5 4 by (rule mp)
qed
qed

lemma ej_6: joslopjim4
assumes "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof (rule impI)
assume "(p⟶q)"
show "p⟶r"
proof (rule impI)
assume "p"
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
p ⊢ q ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_7: (NO SE QUE ESTA MAL)
assumes "p"
shows "q ⟶ p"

proof -

{ assume 2: "q"
have 3: "p" using 1
}
hence 4: "q⟶p" using 2 3 by (rule impI)

show "q⟶p" using 4 by this

oops

lemma ejercicio_7: carmarria, inmbenber
assumes 1: "p"
shows "q ⟶ p"
proof -
{assume 2: "q"
have 3: "p" using 1 by this}
thus "q ⟶ p" by (rule impI)
qed

lemma ejercicio_7:marmedmar3
assumes "p"
shows "q ⟶ p"
proof
assume "q"
show "p" using assms(1) by this
qed

lemma ej_7: joslopjim4
assumes "p"
shows "q ⟶ p"
proof (rule impI)
assume "q"
show "p" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
⊢ p ⟶ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_8: carmarria
"p ⟶ (q ⟶ p)"

proof -
{assume 1: p
{assume 2: q
have 3: p using 1 by this
}
hence 4: "q ⟶ p" by (rule impI)
}
thus "p ⟶ (q ⟶ p)" by (rule impI)
qed

lemma ejercicio_8: marmedmar3
"p ⟶ (q ⟶ p)"
proof
assume 1: "p"
show "(q ⟶ p)"
proof
assume 2: "q"
show "p" using 1 by this
qed
qed

lemma ej_8: joslopjim4
shows "p⟶(q⟶p)"
proof (rule impI)
assume "p"
show "q⟶p"
proof (rule impI)
assume "q"
show "p" using `p` by this
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_9: carmarria
assumes 1: "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof -
{assume 2: "q ⟶ r"
{assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)
}
hence 6: "p ⟶ r" by (rule impI)
}
thus 7: "(q ⟶ r) ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_9: marmedmar3
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof
assume "(q ⟶ r)"
show "(p ⟶ r)"
proof
assume "p"
with `(p ⟶ q)` have "q" by (rule mp)
with `(q ⟶ r)` show "r" by (rule mp)
qed
qed

lemma ej_9: joslopjim4
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof (rule impI)
assume "q⟶r"
show "p⟶r"
proof (rule impI)
assume "p"
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))
------------------------------------------------------------------ *}

lemma ejercicio_10: carmarria
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"

proof-
{assume 2: r
{assume 3: q
{assume 4: p
have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp)
have 6: "r ⟶ s" using 5 3 by (rule mp)
have 7: s using 6 2 by (rule mp)
}
hence 8: "p ⟶ s" by (rule impI)
}
hence 9: "q ⟶ (p ⟶ s)" by (rule impI)
}
thus "r ⟶ (q ⟶ (p ⟶ s))" by (rule impI)
qed

lemma ejercicio_10: marmedmar3
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof
assume 2: "r"
show "(q ⟶ (p ⟶ s))"
proof
assume 3: "q"
show "(p ⟶ s)"
proof
assume 4: "p"
have 5: "(q ⟶ (r ⟶ s))" using 1 4 by (rule mp)
have 6: "(r ⟶ s)" using 5 3 by (rule mp)
show "s" using 6 2 by (rule mp)
qed
qed
qed

lemma ej_10: joslopjim4
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof (rule impI)
assume "r"
show "q⟶(p⟶s)"
proof (rule impI)
assume "q"
show "p⟶s"
proof (rule impI)
assume "p"
have "q ⟶ (r ⟶ s)" using `p ⟶ (q ⟶ (r ⟶ s))` `p` by (rule mp)
have "r⟶s" using `q ⟶ (r ⟶ s)` `q` by (rule mp)
show "s" using `r-->s` `r` by (rule mp)
qed
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))
------------------------------------------------------------------ *}

lemma ejercicio_11: carmarria
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof -
{assume 1: "p ⟶ (q ⟶ r)"
{assume 2: "p ⟶ q"
{assume 3: p
have 4: q using 2 3 by (rule mp)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: r using 5 4 by (rule mp)
}
hence 7: "p ⟶ r" by (rule impI)
}
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)
}
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)
qed

lemma ejercicio_11: marmedmar3
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof
assume 1: "(p ⟶ (q ⟶ r))"
show "((p ⟶ q) ⟶ (p ⟶ r))"
proof
assume 2: "(p ⟶ q)"
show "(p ⟶ r)"
proof
assume 3: "p"
have 4: "(q ⟶ r)" using 1 3 by (rule mp)
have 5: "q" using 2 3 by (rule mp)
show "r" using 4 5 by (rule mp)
qed
qed
qed

lemma ej_11: joslopjim4
shows "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof (rule impI)
assume "p⟶(q⟶r)"
show "(p⟶q)⟶(p⟶r)"
proof (rule impI)
assume "p⟶q"
show "p⟶r"
proof (rule impI)
assume "p"
have "q" using `p⟶q` `p` by (rule mp)
have "q⟶r" using `p⟶(q⟶r)` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_12: carmarria
assumes 1: "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof -
{assume 2: p
{assume 3: q
{assume 4: p
have 5: q using 3 by this
}
hence 6: "p ⟶ q" by (rule impI)
have 7: r using 1 6 by (rule mp)
}
hence 8: "q ⟶ r" by (rule impI)
}
thus "p ⟶ (q ⟶ r)" by (rule impI)
qed

lemma ej_12: joslopjim4
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof (rule impI)
assume "p"
show "q⟶r"
proof (rule impI)
assume "q"
have "p⟶q"
proof (rule impI)
assume "p"
show "q" using `q` by this
qed
show "r" using `(p ⟶ q) ⟶ r` `p⟶q` by (rule mp)
qed
qed

section {* Conjunciones *}

text {* ---------------------------------------------------------------
Ejercicio 13. Demostrar
p, q ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_13: carmarria marmedmar3
assumes "p"
"q"
shows "p ∧ q"

proof-
show "p \<and> q" using assms(1) assms(2) by (rule conjI)
qed

lemma ej_13: joslopjim4
assumes "p"
"q"
shows "p ∧ q"
proof-
show "p∧q" using `p` `q` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
p ∧ q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_14: carmarria marmedmar3
assumes "p ∧ q"
shows "p"

proof -
show "p" using assms by (rule conjunct1)
qed

lemma ej_14: joslopjim4
assumes "p ∧ q"
shows "p"
proof -
show "p" using assms by (rule conjunct1)
qed

text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
p ∧ q ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_15: carmarria marmedmar3
assumes "p ∧ q"
shows "q"

proof -
show q using assms(1) by (rule conjunct2)
qed

lemma ej_15: joslopjim4
assumes "p ∧ q"
shows "q"
proof-
show "q" using assms by (rule conjunct2)
qed

text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_16: carmarria marmedmar3
assumes 1: "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"
proof -
have 2: "q \<and> r" using 1 by (rule conjunct2)
have 3: p using 1 by (rule conjunct1)
have 4: q using 2 by (rule conjunct1)
have 5: r using 2 by (rule conjunct2)
have 6: "p \<and> q" using 3 4 by (rule conjI)
show "(p \<and> q) \<and> r" using 6 5 by (rule conjI)
qed

lemma ej_16: joslopjim4
assumes "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"
proof-
have "p" using assms by (rule conjunct1)
have "q∧r" using assms by (rule conjunct2)
have "q" using `q∧r` by (rule conjunct1)
have "r" using `q∧r` by (rule conjunct2)
have "p∧q" using `p` `q` by (rule conjI)
show "(p∧q)∧r" using `p∧q` `r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 17. Demostrar
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_17:
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof (rule conjI)
have "p ∧ q" using assms ..
thus "p" ..
next
show "q ∧ r"
proof (rule conjI)
have "p ∧ q" using assms ..
thus "q" ..
next
show "r" using assms ..
qed
qed

lemma ejercicio_17: carmarria marmedmar3
assumes 1:"(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof-
have 2:"p \<and> q" using 1 by (rule conjunct1)
have 3: r using 1 by (rule conjunct2)
have 4: p using 2 by (rule conjunct1)
have 5: q using 2 by (rule conjunct2)
have 6: "q \<and> r" using 5 3 by (rule conjI)
show "p \<and> (q \<and> r)" using 4 6 by (rule conjI)
qed

lemma ej_17: joslopjim4
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof-
have "p∧q" using assms by (rule conjunct1)
have "r" using assms by (rule conjunct2)
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
have "q∧r" using `q` `r` by (rule conjI)
show "p∧(q∧r)" using `p` `q∧r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
p ∧ q ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_18: carmarria
assumes "p ∧ q"
shows "p ⟶ q"
proof -
{assume p
have q using assms(1) by (rule conjunct2)
}
thus "p ⟶ q" by (rule impI)
qed

lemma ejercicio_18: marmedmar3
assumes "p ∧ q"
shows "p ⟶ q"
proof
assume "p"
show "q" using assms by (rule conjunct2)
qed

lemma ej_18: joslopjim4
assumes "p ∧ q"
shows "p ⟶ q"
proof (rule impI)
assume "p"
show "q" using assms by (rule conjunct2)
qed

text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_19: carmarria
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof -
{assume 2: p
have 3: "p ⟶ q" using 1 by (rule conjunct1)
have 4: "p ⟶ r" using 1 by (rule conjunct2)
have 5: q using 3 2 by (rule mp)
have 6: r using 4 2 by (rule mp)
have 7: "q \<and> r" using 5 6 by (rule conjI)
}
thus "p ⟶ q \<and> r" by (rule impI)
qed

lemma ejercicio_19: marmedmar3
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof
have 2: "(p ⟶ q)" using assms by (rule conjunct1)
have 3: "(p ⟶ r)" using assms by (rule conjunct2)
assume 4: "p"
with `(p ⟶ q)` have 5: "q" by (rule mp)
have 6: "r" using 3 4 by (rule mp)
show "q ∧ r" using 5 6 by (rule conjI)
qed

lemma ej_19: joslopjim4
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof (rule impI)
assume "p"
have "p⟶q" using assms by (rule conjunct1)
have "q" using `p⟶q` `p` by (rule mp)
have "p⟶r" using assms by (rule conjunct2)
have "r" using `p⟶r` `p` by (rule mp)
show "q∧r" using `q` `r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_20: carmarria
assumes 1: "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof -
{assume 2: p
have 3: "q \<and> r" using 1 2 by (rule mp)
have 4: q using 3 by (rule conjunct1)
}
hence 5: "p ⟶ q" by (rule impI)
{assume 6: p
have 7: "q \<and> r" using 1 6 by (rule mp)
have 8: r using 7 by (rule conjunct2)
}
hence 9: "p ⟶ r" by (rule impI)
show "(p ⟶ q) \<and> (p ⟶ r)" using 5 9 by (rule conjI)
qed

lemma ej_20: joslopjim4
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof-
have "p⟶q"
proof
assume "p"
have "q∧r" using assms `p` by (rule mp)
show "q" using `q∧r` by (rule conjunct1)
qed
have "p⟶r"
proof
assume "p"
have "q∧r" using assms `p` by (rule mp)
show "r" using `q∧r` by (rule conjunct2)
qed
show "(p⟶q)∧(p⟶r)" using `p⟶q` `p⟶r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_21: carmarria
assumes 1: "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof -
{assume 2: "p \<and> q"
have 3: p using 2 by (rule conjunct1)
have 4: q using 2 by (rule conjunct2)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: r using 5 4 by (rule mp)
}
thus "p ∧ q ⟶ r" by (rule impI)
qed

lemma ejercicio_21: marmedmar3
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof
assume "p ∧ q"
hence "p" by (rule conjunct1)
with assms have "(q ⟶ r)" by (rule mp)
have "q" using `p ∧ q` by (rule conjunct2)
with `(q ⟶ r)` show "r" by (rule mp)
qed

lemma ej_21: joslopjim4
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof (rule impI)
assume "p∧q"
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
have "q⟶r" using assms `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_22: carmarria
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof -
{assume p
{assume q
have "p \<and> q" using `p` `q` by (rule conjI)
have r using assms(1) `p \<and> q` by (rule mp)
}
hence "q ⟶ r" by (rule impI)
}
thus "p ⟶ (q ⟶ r)" by (rule impI)
qed

lemma ejercicio_22: marmedmar3
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof
assume "p"
show "(q ⟶ r)"
proof
assume "q"
with `p` have "p ∧ q" by (rule conjI)
with assms show "r" by (rule mp)
qed
qed

lemma ej_22: joslopjim4
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof (rule impI)
assume "p"
show "q⟶r"
proof (rule impI)
assume "q"
have "p∧q" using `p` `q` by (rule conjI)
show "r" using assms `p∧q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_23: carmarria
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof -
{assume "p \<and> q"
{assume p
have q using `p \<and> q` by (rule conjunct2)
}
hence "p ⟶ q" by (rule impI)
have r using assms(1) `p ⟶ q` by (rule mp)
}
thus "p \<and> q ⟶ r" by (rule impI)
qed

lemma ejercicio_23: marmedmar3
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof
assume "p ∧ q"
have "p" using `p ∧ q` by (rule conjunct1)
have "q" using `p ∧ q` by (rule conjunct2)
hence "(p ⟶ q)" by (rule impI)
with assms show "r" by (rule mp)
qed

lemma ej_23: joslopjim4
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof (rule impI)
assume "p∧q"
have "p⟶q"
proof
assume "p"
show "q" using `p∧q` by (rule conjunct2)
qed
show "r" using assms `p⟶q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_24: carmarria
assumes 1: "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof -
{assume 2: "p ⟶ q"
have 3: p using 1 by (rule conjunct1)
have 4: "q ⟶ r" using 1 by (rule conjunct2)
have 5: q using 2 3 by (rule mp)
have 6: r using 4 5 by (rule mp)
}
thus "(p ⟶ q) ⟶ r" by (rule impI)
qed

lemma ejercicio_24: marmedmar3
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof
assume "(p ⟶ q)"
have "(q ⟶ r)" using assms by (rule conjunct2)
have "p" using assms by (rule conjunct1)
with `(p ⟶ q)` have "q" by (rule mp)
with `(q ⟶ r)` show "r" by (rule mp)
qed

lemma ej_24: joslopjim4
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof (rule impI)
assume "p⟶q"
have "p" using assms by (rule conjunct1)
have "q⟶r" using assms by (rule conjunct2)
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

section {* Disyunciones *}

text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
p ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_25: carmarria
assumes "p"
shows "p ∨ q"
proof -
show "p | q" using assms(1) by (rule disjI1)
qed

lemma ej_25: joslopjim4
assumes "p"
shows "p ∨ q"
proof (rule disjI1)
show "p" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
q ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_26: carmarria
assumes "q"
shows "p ∨ q"
proof -
show "p | q" using assms(1) by (rule disjI2)
qed

lemma ej_26: joslopjim4
assumes "q"
shows "p ∨ q"
proof (rule disjI2)
show "q" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar
p ∨ q ⊢ q ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_27: carmarria
assumes "p ∨ q"
shows "q ∨ p"
proof -
have "p | q" using assms(1) by this
moreover
{assume p
have "q | p" using `p` by (rule disjI2)
}
moreover
{assume q
have "q | p" using `q` by (rule disjI1)
}
ultimately show "q | p" by (rule disjE)
qed

lemma ej_27: joslopjim4
assumes "p ∨ q"
shows "q ∨ p"
using assms
proof
assume "p"
then show "q∨p" by(rule disjI2)
next
assume "q"
show "q∨p" using `q` by(rule disjI1)
qed

text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_28: carmarria
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof -
{assume "p | q"
moreover
{assume p
have "p | r" using `p` by (rule disjI1)
}
moreover
{assume q
have r using assms(1) `q` by (rule mp)
have "p | r" using `r` by (rule disjI2)
}
ultimately have "p | r" by (rule disjE)
}
thus "p | q ⟶ p | r" by (rule impI)
qed

lemma ej_28: joslopjim4
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof (rule impI)
assume "p∨q"
show "p∨r"
using `p∨q`
proof
assume "p"
show "p∨r" using `p` by (rule disjI1)
next
assume "q"
have "r" using assms `q` by (rule mp)
show "p∨r" using `r` by (rule disjI2)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 29. Demostrar
p ∨ p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_29: carmarria
assumes "p ∨ p"
shows "p"
proof -
have "p | p" using assms(1) by this
moreover
{assume p
}
moreover
{assume p
}
ultimately show p by (rule disjE)
qed

lemma ej_29: joslopjim4
assumes "p ∨ p"
shows "p"
proof (rule ccontr)
assume "¬p"
show False
using assms
proof
assume "p"
show False using `¬p` `p` by (rule notE)
next
assume "p"
show False using `¬p` `p` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar
p ⊢ p ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_30: carmarria
assumes "p"
shows "p ∨ p"
proof -
show "p | p" using assms(1) by (rule disjI1)
qed

lemma ej_30: joslopjim4
assumes "p"
shows "p ∨ p"
proof (rule disjI1)
show "p" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_31: carmarria
assumes 1: "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
proof -
have "p | (q | r)" using 1 by this
moreover {assume 2: p
have 3: "p | q" using 2 by (rule disjI1)
have 4: "(p | q) | r" using 3 by (rule disjI1)
}
moreover {assume 5: "q | r"
moreover {assume 6: q
have 7: "p | q" using 6 by (rule disjI2)
have 8: "(p | q) | r " using 7 by (rule disjI1)
}
moreover {assume 9: r
have 10: "(p | q) | r " using 9 by (rule disjI2)
}
ultimately have 11: "(p | q) | r " by (rule disjE)
}
ultimately show "(p | q) | r " by (rule disjE)
qed

lemma ej_31: joslopjim4
assumes "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
using assms
proof
assume "p"
have "p∨q" using `p` by (rule disjI1)
thus "(p ∨ q) ∨ r" by (rule disjI1)
next
assume "q∨r"
show "(p ∨ q) ∨ r"
using `q∨r`
proof
assume "q"
have "p∨q" using `q` by (rule disjI2)
show "(p ∨ q) ∨ r" using `p∨q` by (rule disjI1)
next
assume "r"
show "(p ∨ q) ∨ r" using `r` by (rule disjI2)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_32: carmarria
assumes 1: "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
proof -
have "(p | q) | r" using 1 by this
moreover {assume 2: "p | q"
moreover {assume 3: p
have 4: "p | (q | r)" using 3 by (rule disjI1)
}
moreover {assume 5: q
have 6: "q | r" using 5 by (rule disjI1)
have 7: "p | (q | r)" using 6 by (rule disjI2)
}
ultimately have 8: " p | (q | r)" by (rule disjE)
}
moreover {assume 9: r
have 10: "q | r" using 9 by (rule disjI2)
have 11: "p | (q | r)" using 10 by (rule disjI2)
}
ultimately show "p | (q | r)" by (rule disjE)
qed

lemma ej_32: joslopjim4
assumes "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
using assms
proof
assume "p∨q"
show "p ∨ (q ∨ r)"
using `p∨q`
proof
assume "p"
show "p ∨ (q ∨ r)" using `p` by (rule disjI1)
next
assume "q"
have "q∨r" using `q` by (rule disjI1)
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)
qed
next
assume "r"
have "q∨r" using `r` by (rule disjI2)
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)
qed

text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_33: carmarria
assumes 1: "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 2: p using 1 by (rule conjunct1)
have 3: "q | r" using 1 by (rule conjunct2)
moreover {assume 4: q
have 5: "p & q" using 2 4 by (rule conjI)
have 6: "(p & q) | (p & r)" using 5 by (rule disjI1)
}
moreover {assume 7: r
have 8: "p & r" using 2 7 by (rule conjI)
have 9: "(p & q) | (p & r)" using 8 by (rule disjI2)
}
ultimately show "(p & q) | (p & r)" by (rule disjE)
qed

lemma ej_33: joslopjim4
assumes "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof-
have "p" using assms by (rule conjunct1)
have "q∨r" using assms by (rule conjunct2)
show "(p ∧ q) ∨ (p ∧ r)"
using `q∨r`
proof (rule disjE)
assume "q"
have "p∧q" using `p` `q` by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using `p∧q` by (rule disjI1)
next
assume "r"
have "p∧r" using `p` `r` by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using `p∧r` by (rule disjI2)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_34: carmarria
assumes 1: "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
proof -
have "(p & q) | (p & r)" using 1 by this
moreover {assume 2: "p & q"
have 3: p using 2 by (rule conjunct1)
have 4: q using 2 by (rule conjunct2)
have 5: "q | r" using 4 by (rule disjI1)
have 6: "p & (q | r)" using 3 5 by (rule conjI)
}
moreover {assume 7: "p & r"
have 8: p using 7 by (rule conjunct1)
have 9: r using 7 by (rule conjunct2)
have 10: "q | r" using 9 by (rule disjI2)
have 11: "p & (q | r)" using 8 10 by (rule conjI)
}
ultimately show "p & (q | r)" by (rule disjE)
qed

lemma ej_34: joslopjim4
assumes "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
using assms
proof
assume "p∧q"
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
have "q∨r" using `q` by (rule disjI1)
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)
next
assume "p∧r"
have "p" using `p∧r` by (rule conjunct1)
have "r" using `p∧r` by (rule conjunct2)
have "q∨r" using `r` by (rule disjI2)
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_35: carmarria
assumes 1: "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
proof -
have "p | (q & r)" using 1 by this
moreover {assume 2: p
have 3: " p | q" using 2 by (rule disjI1)
have 4: " p | r" using 2 by (rule disjI1)
have 5: "(p | q) & (p | r)" using 3 4 by (rule conjI)
}
moreover {assume 6: "q & r"
have 7: q using 6 by (rule conjunct1)
have 8: r using 6 by (rule conjunct2)
have 9: "p | q" using 7 by (rule disjI2)
have 10: "p | r" using 8 by (rule disjI2)
have 11: "(p | q) & (p | r)" using 9 10 by (rule conjI)
}
ultimately show "(p | q) & (p | r)" by (rule disjE)
qed

lemma ej_35: joslopjim4
assumes "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
using assms
proof (rule disjE)
assume "p"
have "p∨q" using `p` by (rule disjI1)
have "p∨r" using `p` by (rule disjI1)
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)
next
assume "q∧r"
have "q" using `q∧r` by (rule conjunct1)
have "r" using `q∧r` by (rule conjunct2)
have "p∨q" using `q` by (rule disjI2)
have "p∨r" using `r` by (rule disjI2)
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_36: carmarria
assumes 1: "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof -
have 2: "p | q" using 1 by (rule conjunct1)
moreover {assume 3: p
have 4: "p | (q & r)" using 3 by (rule disjI1)
}
moreover {assume 5: q
have 6: "p | r" using 1 by (rule conjunct2)
moreover {assume 7: p
have 8: "p | (q & r)" using 7 by (rule disjI1)
}
moreover {assume 9: r
have 10: "q & r" using 5 9 by (rule conjI)
have 11: "p | (q & r)" using 10 by (rule disjI2)
}
ultimately have 12: "p | (q & r)" by (rule disjE)
}
ultimately show "p | (q & r)" by (rule disjE)
qed

lemma ej_36: joslopjim4
assumes "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof-
have "p∨q" using assms by (rule conjunct1)
have "p∨r" using assms by (rule conjunct2)
show "p∨(q ∧ r)"
using `p∨r`
proof
assume "p"
show "p∨(q ∧ r)" using `p` by (rule disjI1)
next
assume "r"
show "p∨(q ∧ r)"
using `p∨q`
proof
assume "p"
show "p∨(q ∧ r)" using `p` by (rule disjI1)
next
assume "q"
have "q∧r" using `q` `r` by (rule conjI)
show "p∨(q ∧ r)" using `q∧r` by (rule disjI2)
qed
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_37: carmarria
assumes 1: "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof -
have 2: "p ⟶ r" using 1 by (rule conjunct1)
have 3: "q ⟶ r" using 1 by (rule conjunct2)
{assume 4: "p | q"
moreover {assume 5: p
have 6: r using 2 5 by (rule mp)
}
moreover {assume 7: q
have 8: r using 3 7 by (rule mp)
}
ultimately have r by (rule disjE)
}
thus "p | q ⟶ r" by (rule impI)
qed

lemma ej_37: joslopjim4
assumes "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof (rule impI)
assume "p∨q"
have "p⟶r" using assms by (rule conjunct1)
have "q⟶r" using assms by (rule conjunct2)
show "r"
using `p∨q`
proof
assume "p"
show "r" using `p⟶r` `p` by (rule mp)
next
assume "q"
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 38. Demostrar
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_38: carmarria
assumes 1: "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof -
{assume 2: p
have 3: "p | q" using 2 by (rule disjI1)
have 4: r using 1 3 by (rule mp)
}
hence 5: "p ⟶ r" by (rule impI)
{assume 6: q
have 7: "p | q" using 6 by (rule disjI2)
have 8: r using 1 7 by (rule mp)
}
hence 9: "q ⟶ r" by (rule impI)
show "(p ⟶ r) & (q ⟶ r)" using 5 9 by (rule conjI)
qed

lemma ej_38: joslopjim4
assumes "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof-
have "p⟶r"
proof (rule impI)
assume "p"
have "p∨q" using `p` by (rule disjI1)
show "r" using assms `p∨q` by (rule mp)
qed
have "q⟶r"
proof (rule impI)
assume "q"
have "p∨q" using `q` by (rule disjI2)
show "r" using assms `p∨q` by (rule mp)
qed
show "(p ⟶ r) ∧ (q ⟶ r)" using `p⟶r` `q⟶r` by (rule conjI)
qed

section {* Negaciones *}

text {* ---------------------------------------------------------------
Ejercicio 39. Demostrar
p ⊢ ¬¬p
------------------------------------------------------------------ *}

lemma ejercicio_39: carmarria
assumes "p"
shows "¬¬p"
proof -
show "~~p" using assms(1) by (rule notnotI)
qed

lemma ej_39: joslopjim4
assumes "p"
shows "¬¬p"
proof-
show "¬¬p" using assms by (rule notnotI)
qed

text {* ---------------------------------------------------------------
Ejercicio 40. Demostrar
¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_40: carmarria
assumes "¬p"
shows "p ⟶ q"
proof -
{assume p
have "~p" using assms(1) by this
have False using `~p` `p` by (rule notE)
have q using `False` by (rule FalseE)
}
thus "p ⟶ q" by (rule impI)
qed

lemma ej_40: joslopjim4
assumes "¬p"
shows "p ⟶ q"
proof (rule impI)
assume "p"
show "q" using assms `p` by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 41. Demostrar
p ⟶ q ⊢ ¬q ⟶ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_41: carmarria
assumes 1: "p ⟶ q"
shows "¬q ⟶ ¬p"
proof -
{assume 2: "~q"
have 3: "~p" using 1 2 by (rule mt)
}
thus 4: "~q ⟶ ~p" by (rule impI)
qed

lemma ej_41: joslopjim4
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
proof (rule impI)
assume "¬q"
show "¬p" using assms `¬q` by (rule mt)
qed

text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p∨q, ¬q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_42: carmarria
assumes 1: "p∨q" and
2: "¬q"
shows "p"
proof -
have " p | q" using 1 by this
moreover {assume 3: p
}
moreover {assume 4: q
have 5: False using 2 4 by (rule notE)
have 6: p using 5 by (rule FalseE)
}
ultimately show p by (rule disjE)
qed

lemma ej_42: joslopjim4
assumes "p∨q"
"¬q"
shows "p"
using assms(1)
proof
assume "p"
show "p" using `p` by this
next
assume "q"
show "p" using assms(2) `q` by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 43. Demostrar
p ∨ q, ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_43: carmarria
assumes 1: "p ∨ q" and
2: "¬p"
shows "q"
proof -
have " p | q" using 1 by this
moreover {assume 3: p
have 4: False using 2 3 by (rule notE)
have 5: q using 4 by (rule FalseE)
}
moreover {assume 6: q
}
ultimately show q by (rule disjE)
qed

lemma ej_43: joslopjim4
assumes "p ∨ q"
"¬p"
shows "q"
using assms(1)
proof
assume "p"
show "q" using assms(2) `p` by (rule notE)
next
assume "q"
show "q" using `q` by this
qed

text {* ---------------------------------------------------------------
Ejercicio 44. Demostrar
p ∨ q ⊢ ¬(¬p ∧ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_44: carmarria
assumes 1: "p ∨ q"
shows "¬(¬p ∧ ¬q)"
proof -
have "p | q" using 1 by this
moreover {assume 2: p
{assume 3:"~p & ~q"
have 4: "~p" using 3 by (rule conjunct1)
have 5: False using 4 2 by (rule notE)
}
hence 6: "~(~p & ~q)" by (rule notI)
}
moreover {assume 7: q
{assume 8: "~p & ~q"
have 9: "~q" using 8 by (rule conjunct2)
have 10: False using 9 7 by (rule notE)
}
hence 11: "~(~p & ~q)" by (rule notI)
}
ultimately show "~(~p & ~q)" by (rule disjE)
qed

lemma ej_44: joslopjim4
assumes "p∨q"
shows "¬(¬p∧¬q)"
proof (rule notI)
assume "¬p∧¬q"
show False
using assms
proof
assume "p"
have "¬p" using `¬p∧¬q` by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "q"
have "¬q" using `¬p∧¬q` by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 45. Demostrar
p ∧ q ⊢ ¬(¬p ∨ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_45: carmarria
assumes 1: "p ∧ q"
shows "¬(¬p ∨ ¬q)"
proof -
{assume 2: "~p | ~q"
moreover {assume 3: "~p"
have 4: p using 1 by (rule conjunct1)
have 5: False using 3 4 by (rule notE)
}
moreover {assume 6: "~q"
have 7: q using 1 by (rule conjunct2)
have 8: False using 6 7 by (rule notE)
}
ultimately have False by (rule disjE)
}
thus "~(~p | ~q)" by (rule notI)
qed

lemma ej_45: joslopjim4
assumes "p∧q"
shows "¬(¬p∨¬q)"
proof (rule notI)
assume "¬p∨¬q"
then show False
proof (rule disjE)
assume "¬p"
have "p" using assms by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "¬q"
have "q" using assms by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 46. Demostrar
¬(p ∨ q) ⊢ ¬p ∧ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_46: carmarria
assumes 1: "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof -
{assume 2: p
have 3: "p | q" using 2 by (rule disjI1)
have 4: False using 1 3 by (rule notE)
}
hence 5: "~p" by (rule notI)
{assume 6: q
have 7: "p | q" using 6 by (rule disjI2)
have 8: False using 1 7 by (rule notE)
}
hence 9: "~q" by (rule notI)
show "~p & ~q" using 5 9 by (rule conjI)
qed

lemma ej_46: joslopjim4
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof (rule conjI)
show "¬p"
proof (rule ccontr)
assume "¬¬p"
have "p" using `¬¬p` by (rule notnotD)
have "p∨q" using `p` by (rule disjI1)
show False using assms `p∨q` by (rule notE)
qed
show "¬q"
proof (rule ccontr)
assume "¬¬q"
have "q" using `¬¬q` by (rule notnotD)
have "p∨q" using `q` by (rule disjI2)
show False using assms `p∨q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 47. Demostrar
¬p ∧ ¬q ⊢ ¬(p ∨ q)
------------------------------------------------------------------ *}

lemma ejercicio_47: carmarria
assumes 1: "¬p ∧ ¬q"
shows "¬(p ∨ q)"
proof -
{assume 2: "p | q"
moreover {assume 3: p
have 4: "~p" using 1 by (rule conjunct1)
have 5: False using 4 3 by (rule notE)
}
moreover {assume 6: q
have 7: "~q" using 1 by (rule conjunct2)
have 8: False using 7 6 by (rule notE)
}
ultimately have 9: False by (rule disjE)
}
thus "~(p | q)" by (rule notI)
qed

lemma ej_47: joslopjim4
assumes "¬p∧¬q"
shows "¬(p∨q)"
proof (rule notI)
assume "p∨q"
then show False
proof
assume "p"
have "¬p" using assms by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "q"
have "¬q" using assms by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 48. Demostrar
¬p ∨ ¬q ⊢ ¬(p ∧ q)
------------------------------------------------------------------ *}

lemma ejercicio_48: carmarria
assumes 1: "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof -
have "~p | ~q" using 1 by this
moreover {assume 2: "~p"
{assume 3: "p & q"
have 4: p using 3 by (rule conjunct1)
have 5: False using 2 4 by (rule notE)
}
hence 6: "~(p & q)" by (rule notI)
}
moreover {assume 7: "~q"
{assume 8: "p & q"
have 9: q using 8 by (rule conjunct2)
have 10: False using 7 9 by (rule notE)
}
hence 11: "~(p & q)" by (rule notI)
}
ultimately show "~(p & q)" by (rule disjE)
qed

lemma ej_48: joslopjim4
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof (rule notI)
assume "p∧q"
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
show False
using assms
proof
assume "¬p"
have "p" using `p∧q` by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "¬q"
have "q" using `p∧q` by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 49. Demostrar
⊢ ¬(p ∧ ¬p)
------------------------------------------------------------------ *}

lemma ejercicio_49: carmarria
"¬(p ∧ ¬p)"
proof -
{assume 1: "p & ~p"
have 2: p using 1 by (rule conjunct1)
have 3: "~p" using 1 by (rule conjunct2)
have 4: False using 3 2 by (rule notE)
}
thus "~(p & ~p)" by (rule notI)
qed

lemma ej_49: joslopjim4
shows "¬(p∧¬p)"
proof
assume "p∧¬p"
have "p" using `p∧¬p` by (rule conjunct1)
have "¬p" using `p∧¬p` by (rule conjunct2)
show False using `¬p` `p` by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 50. Demostrar
p ∧ ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_50: carmarria
assumes 1: "p ∧ ¬p"
shows "q"
proof -
have 2: p using 1 by (rule conjunct1)
have 3: "~p" using 1 by (rule conjunct2)
have 4: False using 3 2 by (rule notE)
show q using 4 by (rule FalseE)
qed

lemma ej_50: joslopjim4
assumes "p∧¬p"
shows "q"
proof-
have "p" using `p∧¬p` by (rule conjunct1)
have "¬p" using `p∧¬p` by (rule conjunct2)
have False using `¬p` `p` by (rule notE)
then show "q" by (rule FalseE)
qed

text {* ---------------------------------------------------------------
Ejercicio 51. Demostrar
¬¬p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_51: carmarria joslopjim4
assumes "¬¬p"
shows "p"
proof -
show p using assms by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 52. Demostrar
⊢ p ∨ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_52: carmarria
"p ∨ ¬p"
proof -
{assume 1: "~(p | ~p)"
{assume 2: p
have 3: "p | ~p" using 2 by (rule disjI1)
have 4: False using 1 3 by (rule notE)
}
hence 5: "~p" by (rule notI)
have 6: "p | ~p" using 5 by (rule disjI2)
have 7: False using 1 6 by (rule notE)
}
hence 8: "~~(p | ~p)" by (rule notI)
show "p | ~p" using 8 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 53. Demostrar
⊢ ((p ⟶ q) ⟶ p) ⟶ p
------------------------------------------------------------------ *}

lemma ej_53: joslopjim4
shows "((p ⟶ q) ⟶ p) ⟶ p"
proof (rule impI)
assume "(p ⟶ q) ⟶ p"
show "p"
proof (rule ccontr)
assume "¬p"
have "¬(p⟶q)" using `(p ⟶ q) ⟶ p` `¬p` by (rule mt)
have "p⟶q"
proof (rule impI)
assume "p"
show "q" using `¬p` `p` by (rule notE)
qed
show False using `¬(p⟶q)` `p⟶q` by (rule notE)
qed
qed

lemma ejercicio_53: carmarria
"((p ⟶ q) ⟶ p) ⟶ p"
proof -
{assume 1: "(p ⟶ q) ⟶ p"
{assume 2:"~p"
have 3: "~(p ⟶q)" using 1 2 by (rule mt)
{assume 4: p
have 5: False using 2 4 by (rule notE)
have 6: q using 5 by (rule FalseE)
}
hence 7: "p ⟶q" by (rule impI)
have 8: False using 3 7 by (rule notE)
}
hence 9: "~~p" by (rule notI)
have 10: p using 9 by (rule notnotD)
}
thus "((p ⟶ q)⟶p)⟶p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 54. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ej_54: joslopjim4
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof (rule impI)
assume "p"
have "¬¬p" using `p` by (rule notnotI)
have "¬¬q" using assms `¬¬p` by (rule mt)
show "q" using `¬¬q` by (rule notnotD)
qed

lemma ejercicio_54: carmarria
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{assume 2: p
have 3: "~~p" using 2 by (rule notnotI)
have 4: "~~q" using 1 3 by (rule mt)
have 5: q using 4 by (rule notnotD)
}
thus "p ⟶q" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 55. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_55: carmarria
assumes 1: "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
{assume 2: "~(p | q)"
{assume 3: "~p"
{assume 4: "~q"
have 5: "~p & ~q" using 3 4 by (rule conjI)
have 6: False using 1 5 by (rule notE)
}
hence 7: "~~q" by (rule notI)
have 8: q using 7 by (rule notnotD)
have 9: "p | q" using 8 by (rule disjI2)
have 10: False using 2 9 by (rule notE)
}
hence 11: "~~p" by (rule notI)
have 12: p using 11 by(rule notnotD)
have 13: "p | q" using 12 by (rule disjI1)
have 14: False using 2 13 by (rule notE)
}
hence 15: "~~(p | q)" by (rule notI)
show "p | q" using 15 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 56. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_56: carmarria
assumes 1: "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof -
{assume 2: "~p"
have 3: "~p | ~q" using 2 by (rule disjI1)
have 4: False using 1 3 by (rule notE)
}
hence 5: "~~p" by (rule notI)
have 6: p using 5 by (rule notnotD)
{assume 7: "~q"
have 8: "~p | ~q" using 7 by (rule disjI2)
have 9: False using 1 8 by (rule notE)
}
hence 10: "~~q" by (rule notI)
have 11: q using 10 by (rule notnotD)
show 12: " p & q" using 6 11 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 57. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_57: carmarria
assumes 1: "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof -
{assume 2: "~(~p | ~q)"
{assume 3: p
{assume 4: q
have 5: "p & q" using 3 4 by (rule conjI)
have 6: False using 1 5 by (rule notE)
}
hence 7: "~q" by (rule notI)
have 8: "~p | ~q" using 7 by (rule disjI2)
have 9: False using 2 8 by (rule notE)
}
hence 10: "~p" by (rule notI)
have 11: "~p | ~q" using 10 by (rule disjI1)
have 12: False using 2 11 by (rule notE)
}
hence 13: "~~(~p | ~q)" by (rule notI)
show "~p | ~q" using 13 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 58. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_58:
"(p ⟶ q) ∨ (q ⟶ p)"
oops

end

</source>

Relación 4

2018-03-22T14:18:55Z

Carmarria:

<source lang = "isar">

chapter {* R4: Deducción natural proposicional *}

theory R4
imports Main
begin

text {*
---------------------------------------------------------------------
El objetivo de esta relación es demostrar cada uno de los ejercicios
usando sólo las reglas básicas de deducción natural de la lógica
proposicional (sin usar el método auto).

Las reglas básicas de la deducción natural son las siguientes:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· notnotI: P ⟹ ¬¬ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P
---------------------------------------------------------------------
*}

text {*
Se usarán las reglas notnotI y mt que demostramos a continuación. *}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

section {* Implicaciones *}

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
p ⟶ q, p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_1: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ q" and
2: "p"
shows "q"

proof -
show 1: "q" using 1 2 by (rule mp)
qed

lemma ej_1: joslopjim4
assumes "p --> q"
"p"
shows "q"
proof -
show "q" using assms(1) assms(2) by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
p ⟶ q, q ⟶ r, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_2: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ q" and
2: "q ⟶ r" and
3: "p"
shows "r"

proof-
have 4: "q" using 1 3 by (rule mp)
show "r" using 2 4 by (rule mp)
qed

lemma ej_2: joslopjim4
assumes "p ⟶ q"
"q ⟶ r"
"p"
shows "r"
proof -
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_3: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ (q ⟶ r)" and
2: "p ⟶ q" and
3: "p"
shows "r"

proof-
have 4: "q" using 2 3 by (rule mp)
have 5: "(q ⟶ r)" using 1 3 by (rule mp)
show "r" using 5 4 by (rule mp)

qed

lemma ej_3: joslopjim4
assumes "p ⟶ (q ⟶ r)"
"p ⟶ q"
"p"
shows "r"
proof -
have "q" using `p⟶q` `p` by (rule mp)
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
p ⟶ q, q ⟶ r ⊢ p ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_4: josrodjim2
assumes 1: "p ⟶ q" and
2: "q ⟶ r"
shows "p ⟶ r"

proof-
{ assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)}
hence 6: "p ⟶ r" using 3 5 by (rule impI)
show "p⟶r" using 6 by this

oops

lemma ejercicio_4: carmarria
assumes 1: "p ⟶ q" and
2: "q ⟶ r"
shows "p ⟶ r"
proof -
{ assume 3: p
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)
}
thus "p ⟶ r" by (rule impI)
qed
lemma ejercicio_4: inmbenber
assumes 1: "p ⟶ q" and
2: "q ⟶ r"
shows "p ⟶ r"
proof -
{ assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp) }
thus "p ⟶ r" by (rule impI)
qed

lemma ejercicio_4:marmedmar3
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof
assume "p"
with assms(1) have "q" by (rule mp)
with assms(2) show "r" by (rule mp)
qed

lemma ej_4: joslopjim4
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof (rule impI)
assume "p"
have "q" using assms(1) `p` by (rule mp)
show "r" using assms(2) `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_5: josrodjim2 (NO DA ERROR, PERO SI ALERTA)
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

proof -

{ assume 2: "q"
{assume 3: "p"
have 4: "q ⟶r" using 1 3 by (rule mp)
have 5: "r" using 4 2 by (rule mp)}
hence 6: "p⟶r" using 3 5 by (rule impI)}
hence 7: "q ⟶ (p ⟶ r)" using 2 6 by (rule impI)
show "q ⟶ (p ⟶ r)" using 7 by this

oops

lemma ejercicio_5: carmarria
assumes 1: "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

proof -
{assume 2: q
{assume 3: p
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "r" using 4 2 by (rule mp)
}
hence " p ⟶ r" by (rule impI)
}
thus "q ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_5: inmbenber
assumes 1: "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

proof -
{assume 2: "q"
{assume 3: "p"
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "r" using 4 2 by (rule mp) }
hence " p ⟶ r" by (rule impI) }
thus "q ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_5: marmedmar3
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"
proof
assume "q"
show "(p ⟶ r)"
proof
assume "p"
with assms(1) have "(q ⟶ r)" by (rule mp)
thus "r" using `q` by (rule mp)
qed
qed

lemma ej_5: joslopjim4
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"
proof (rule impI)
assume "q"
show "p⟶r"
proof (rule impI)
assume "p"
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 6. Demostrar
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_6:josrodjim2
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"

proof-

{assume 2: "p⟶q"
{assume 3: "p"
have 4: "q⟶r" using 1 3 by (rule mp)
have 5: "q" using 2 3 by (rule mp)
have 6: "r" using 4 5 by (rule mp)}
hence 7: "p⟶r" using 3 6 by (rule impI)}
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" using 2 7 by (rule impI)
show "(p ⟶ q) ⟶ (p ⟶ r)" using 8 by this
oops

lemma ejercicio_6: carmarria, inmbenber
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof -
{assume 2: "p ⟶ q"
{assume 3: "p"
have 4: "q" using 2 3 by (rule mp)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: "r" using 5 4 by (rule mp)
}
hence 7: "p ⟶ r" by (rule impI)
}
thus "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_6:marmedmar3
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof
assume 2: "(p ⟶ q)"
show "(p ⟶ r)"
proof
assume 3: "p"
have 4: "q" using 2 3 by (rule mp)
have 5: "(q ⟶ r)" using 1 3 by (rule mp)
show "r" using 5 4 by (rule mp)
qed
qed

lemma ej_6: joslopjim4
assumes "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof (rule impI)
assume "(p⟶q)"
show "p⟶r"
proof (rule impI)
assume "p"
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
p ⊢ q ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_7: (NO SE QUE ESTA MAL)
assumes "p"
shows "q ⟶ p"

proof -

{ assume 2: "q"
have 3: "p" using 1
}
hence 4: "q⟶p" using 2 3 by (rule impI)

show "q⟶p" using 4 by this

oops

lemma ejercicio_7: carmarria, inmbenber
assumes 1: "p"
shows "q ⟶ p"
proof -
{assume 2: "q"
have 3: "p" using 1 by this}
thus "q ⟶ p" by (rule impI)
qed

lemma ejercicio_7:marmedmar3
assumes "p"
shows "q ⟶ p"
proof
assume "q"
show "p" using assms(1) by this
qed

lemma ej_7: joslopjim4
assumes "p"
shows "q ⟶ p"
proof (rule impI)
assume "q"
show "p" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
⊢ p ⟶ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_8: carmarria
"p ⟶ (q ⟶ p)"

proof -
{assume 1: p
{assume 2: q
have 3: p using 1 by this
}
hence 4: "q ⟶ p" by (rule impI)
}
thus "p ⟶ (q ⟶ p)" by (rule impI)
qed

lemma ejercicio_8: marmedmar3
"p ⟶ (q ⟶ p)"
proof
assume 1: "p"
show "(q ⟶ p)"
proof
assume 2: "q"
show "p" using 1 by this
qed
qed

lemma ej_8: joslopjim4
shows "p⟶(q⟶p)"
proof (rule impI)
assume "p"
show "q⟶p"
proof (rule impI)
assume "q"
show "p" using `p` by this
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_9: carmarria
assumes 1: "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof -
{assume 2: "q ⟶ r"
{assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)
}
hence 6: "p ⟶ r" by (rule impI)
}
thus 7: "(q ⟶ r) ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_9: marmedmar3
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof
assume "(q ⟶ r)"
show "(p ⟶ r)"
proof
assume "p"
with `(p ⟶ q)` have "q" by (rule mp)
with `(q ⟶ r)` show "r" by (rule mp)
qed
qed

lemma ej_9: joslopjim4
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof (rule impI)
assume "q⟶r"
show "p⟶r"
proof (rule impI)
assume "p"
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))
------------------------------------------------------------------ *}

lemma ejercicio_10: carmarria
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"

proof-
{assume 2: r
{assume 3: q
{assume 4: p
have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp)
have 6: "r ⟶ s" using 5 3 by (rule mp)
have 7: s using 6 2 by (rule mp)
}
hence 8: "p ⟶ s" by (rule impI)
}
hence 9: "q ⟶ (p ⟶ s)" by (rule impI)
}
thus "r ⟶ (q ⟶ (p ⟶ s))" by (rule impI)
qed

lemma ejercicio_10: marmedmar3
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof
assume 2: "r"
show "(q ⟶ (p ⟶ s))"
proof
assume 3: "q"
show "(p ⟶ s)"
proof
assume 4: "p"
have 5: "(q ⟶ (r ⟶ s))" using 1 4 by (rule mp)
have 6: "(r ⟶ s)" using 5 3 by (rule mp)
show "s" using 6 2 by (rule mp)
qed
qed
qed

lemma ej_10: joslopjim4
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof (rule impI)
assume "r"
show "q⟶(p⟶s)"
proof (rule impI)
assume "q"
show "p⟶s"
proof (rule impI)
assume "p"
have "q ⟶ (r ⟶ s)" using `p ⟶ (q ⟶ (r ⟶ s))` `p` by (rule mp)
have "r⟶s" using `q ⟶ (r ⟶ s)` `q` by (rule mp)
show "s" using `r-->s` `r` by (rule mp)
qed
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))
------------------------------------------------------------------ *}

lemma ejercicio_11: carmarria
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof -
{assume 1: "p ⟶ (q ⟶ r)"
{assume 2: "p ⟶ q"
{assume 3: p
have 4: q using 2 3 by (rule mp)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: r using 5 4 by (rule mp)
}
hence 7: "p ⟶ r" by (rule impI)
}
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)
}
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)
qed

lemma ejercicio_11: marmedmar3
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof
assume 1: "(p ⟶ (q ⟶ r))"
show "((p ⟶ q) ⟶ (p ⟶ r))"
proof
assume 2: "(p ⟶ q)"
show "(p ⟶ r)"
proof
assume 3: "p"
have 4: "(q ⟶ r)" using 1 3 by (rule mp)
have 5: "q" using 2 3 by (rule mp)
show "r" using 4 5 by (rule mp)
qed
qed
qed

lemma ej_11: joslopjim4
shows "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof (rule impI)
assume "p⟶(q⟶r)"
show "(p⟶q)⟶(p⟶r)"
proof (rule impI)
assume "p⟶q"
show "p⟶r"
proof (rule impI)
assume "p"
have "q" using `p⟶q` `p` by (rule mp)
have "q⟶r" using `p⟶(q⟶r)` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_12: carmarria
assumes 1: "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof -
{assume 2: p
{assume 3: q
{assume 4: p
have 5: q using 3 by this
}
hence 6: "p ⟶ q" by (rule impI)
have 7: r using 1 6 by (rule mp)
}
hence 8: "q ⟶ r" by (rule impI)
}
thus "p ⟶ (q ⟶ r)" by (rule impI)
qed

lemma ej_12: joslopjim4
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof (rule impI)
assume "p"
show "q⟶r"
proof (rule impI)
assume "q"
have "p⟶q"
proof (rule impI)
assume "p"
show "q" using `q` by this
qed
show "r" using `(p ⟶ q) ⟶ r` `p⟶q` by (rule mp)
qed
qed

section {* Conjunciones *}

text {* ---------------------------------------------------------------
Ejercicio 13. Demostrar
p, q ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_13: carmarria marmedmar3
assumes "p"
"q"
shows "p ∧ q"

proof-
show "p \<and> q" using assms(1) assms(2) by (rule conjI)
qed

lemma ej_13: joslopjim4
assumes "p"
"q"
shows "p ∧ q"
proof-
show "p∧q" using `p` `q` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
p ∧ q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_14: carmarria marmedmar3
assumes "p ∧ q"
shows "p"

proof -
show "p" using assms by (rule conjunct1)
qed

lemma ej_14: joslopjim4
assumes "p ∧ q"
shows "p"
proof -
show "p" using assms by (rule conjunct1)
qed

text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
p ∧ q ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_15: carmarria marmedmar3
assumes "p ∧ q"
shows "q"

proof -
show q using assms(1) by (rule conjunct2)
qed

lemma ej_15: joslopjim4
assumes "p ∧ q"
shows "q"
proof-
show "q" using assms by (rule conjunct2)
qed

text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_16: carmarria marmedmar3
assumes 1: "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"
proof -
have 2: "q \<and> r" using 1 by (rule conjunct2)
have 3: p using 1 by (rule conjunct1)
have 4: q using 2 by (rule conjunct1)
have 5: r using 2 by (rule conjunct2)
have 6: "p \<and> q" using 3 4 by (rule conjI)
show "(p \<and> q) \<and> r" using 6 5 by (rule conjI)
qed

lemma ej_16: joslopjim4
assumes "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"
proof-
have "p" using assms by (rule conjunct1)
have "q∧r" using assms by (rule conjunct2)
have "q" using `q∧r` by (rule conjunct1)
have "r" using `q∧r` by (rule conjunct2)
have "p∧q" using `p` `q` by (rule conjI)
show "(p∧q)∧r" using `p∧q` `r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 17. Demostrar
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_17:
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof (rule conjI)
have "p ∧ q" using assms ..
thus "p" ..
next
show "q ∧ r"
proof (rule conjI)
have "p ∧ q" using assms ..
thus "q" ..
next
show "r" using assms ..
qed
qed

lemma ejercicio_17: carmarria marmedmar3
assumes 1:"(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof-
have 2:"p \<and> q" using 1 by (rule conjunct1)
have 3: r using 1 by (rule conjunct2)
have 4: p using 2 by (rule conjunct1)
have 5: q using 2 by (rule conjunct2)
have 6: "q \<and> r" using 5 3 by (rule conjI)
show "p \<and> (q \<and> r)" using 4 6 by (rule conjI)
qed

lemma ej_17: joslopjim4
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof-
have "p∧q" using assms by (rule conjunct1)
have "r" using assms by (rule conjunct2)
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
have "q∧r" using `q` `r` by (rule conjI)
show "p∧(q∧r)" using `p` `q∧r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
p ∧ q ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_18: carmarria
assumes "p ∧ q"
shows "p ⟶ q"
proof -
{assume p
have q using assms(1) by (rule conjunct2)
}
thus "p ⟶ q" by (rule impI)
qed

lemma ejercicio_18: marmedmar3
assumes "p ∧ q"
shows "p ⟶ q"
proof
assume "p"
show "q" using assms by (rule conjunct2)
qed

lemma ej_18: joslopjim4
assumes "p ∧ q"
shows "p ⟶ q"
proof (rule impI)
assume "p"
show "q" using assms by (rule conjunct2)
qed

text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_19: carmarria
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof -
{assume 2: p
have 3: "p ⟶ q" using 1 by (rule conjunct1)
have 4: "p ⟶ r" using 1 by (rule conjunct2)
have 5: q using 3 2 by (rule mp)
have 6: r using 4 2 by (rule mp)
have 7: "q \<and> r" using 5 6 by (rule conjI)
}
thus "p ⟶ q \<and> r" by (rule impI)
qed

lemma ejercicio_19: marmedmar3
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof
have 2: "(p ⟶ q)" using assms by (rule conjunct1)
have 3: "(p ⟶ r)" using assms by (rule conjunct2)
assume 4: "p"
with `(p ⟶ q)` have 5: "q" by (rule mp)
have 6: "r" using 3 4 by (rule mp)
show "q ∧ r" using 5 6 by (rule conjI)
qed

lemma ej_19: joslopjim4
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof (rule impI)
assume "p"
have "p⟶q" using assms by (rule conjunct1)
have "q" using `p⟶q` `p` by (rule mp)
have "p⟶r" using assms by (rule conjunct2)
have "r" using `p⟶r` `p` by (rule mp)
show "q∧r" using `q` `r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_20: carmarria
assumes 1: "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof -
{assume 2: p
have 3: "q \<and> r" using 1 2 by (rule mp)
have 4: q using 3 by (rule conjunct1)
}
hence 5: "p ⟶ q" by (rule impI)
{assume 6: p
have 7: "q \<and> r" using 1 6 by (rule mp)
have 8: r using 7 by (rule conjunct2)
}
hence 9: "p ⟶ r" by (rule impI)
show "(p ⟶ q) \<and> (p ⟶ r)" using 5 9 by (rule conjI)
qed

lemma ej_20: joslopjim4
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof-
have "p⟶q"
proof
assume "p"
have "q∧r" using assms `p` by (rule mp)
show "q" using `q∧r` by (rule conjunct1)
qed
have "p⟶r"
proof
assume "p"
have "q∧r" using assms `p` by (rule mp)
show "r" using `q∧r` by (rule conjunct2)
qed
show "(p⟶q)∧(p⟶r)" using `p⟶q` `p⟶r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_21: carmarria
assumes 1: "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof -
{assume 2: "p \<and> q"
have 3: p using 2 by (rule conjunct1)
have 4: q using 2 by (rule conjunct2)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: r using 5 4 by (rule mp)
}
thus "p ∧ q ⟶ r" by (rule impI)
qed

lemma ejercicio_21: marmedmar3
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof
assume "p ∧ q"
hence "p" by (rule conjunct1)
with assms have "(q ⟶ r)" by (rule mp)
have "q" using `p ∧ q` by (rule conjunct2)
with `(q ⟶ r)` show "r" by (rule mp)
qed

lemma ej_21: joslopjim4
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof (rule impI)
assume "p∧q"
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
have "q⟶r" using assms `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_22: carmarria
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof -
{assume p
{assume q
have "p \<and> q" using `p` `q` by (rule conjI)
have r using assms(1) `p \<and> q` by (rule mp)
}
hence "q ⟶ r" by (rule impI)
}
thus "p ⟶ (q ⟶ r)" by (rule impI)
qed

lemma ejercicio_22: marmedmar3
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof
assume "p"
show "(q ⟶ r)"
proof
assume "q"
with `p` have "p ∧ q" by (rule conjI)
with assms show "r" by (rule mp)
qed
qed

lemma ej_22: joslopjim4
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof (rule impI)
assume "p"
show "q⟶r"
proof (rule impI)
assume "q"
have "p∧q" using `p` `q` by (rule conjI)
show "r" using assms `p∧q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_23: carmarria
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof -
{assume "p \<and> q"
{assume p
have q using `p \<and> q` by (rule conjunct2)
}
hence "p ⟶ q" by (rule impI)
have r using assms(1) `p ⟶ q` by (rule mp)
}
thus "p \<and> q ⟶ r" by (rule impI)
qed

lemma ejercicio_23: marmedmar3
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof
assume "p ∧ q"
have "p" using `p ∧ q` by (rule conjunct1)
have "q" using `p ∧ q` by (rule conjunct2)
hence "(p ⟶ q)" by (rule impI)
with assms show "r" by (rule mp)
qed

lemma ej_23: joslopjim4
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof (rule impI)
assume "p∧q"
have "p⟶q"
proof
assume "p"
show "q" using `p∧q` by (rule conjunct2)
qed
show "r" using assms `p⟶q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_24: carmarria
assumes 1: "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof -
{assume 2: "p ⟶ q"
have 3: p using 1 by (rule conjunct1)
have 4: "q ⟶ r" using 1 by (rule conjunct2)
have 5: q using 2 3 by (rule mp)
have 6: r using 4 5 by (rule mp)
}
thus "(p ⟶ q) ⟶ r" by (rule impI)
qed

lemma ejercicio_24: marmedmar3
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof
assume "(p ⟶ q)"
have "(q ⟶ r)" using assms by (rule conjunct2)
have "p" using assms by (rule conjunct1)
with `(p ⟶ q)` have "q" by (rule mp)
with `(q ⟶ r)` show "r" by (rule mp)
qed

lemma ej_24: joslopjim4
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof (rule impI)
assume "p⟶q"
have "p" using assms by (rule conjunct1)
have "q⟶r" using assms by (rule conjunct2)
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

section {* Disyunciones *}

text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
p ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_25: carmarria
assumes "p"
shows "p ∨ q"
proof -
show "p | q" using assms(1) by (rule disjI1)
qed

lemma ej_25: joslopjim4
assumes "p"
shows "p ∨ q"
proof (rule disjI1)
show "p" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
q ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_26: carmarria
assumes "q"
shows "p ∨ q"
proof -
show "p | q" using assms(1) by (rule disjI2)
qed

lemma ej_26: joslopjim4
assumes "q"
shows "p ∨ q"
proof (rule disjI2)
show "q" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar
p ∨ q ⊢ q ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_27: carmarria
assumes "p ∨ q"
shows "q ∨ p"
proof -
have "p | q" using assms(1) by this
moreover
{assume p
have "q | p" using `p` by (rule disjI2)
}
moreover
{assume q
have "q | p" using `q` by (rule disjI1)
}
ultimately show "q | p" by (rule disjE)
qed

lemma ej_27: joslopjim4
assumes "p ∨ q"
shows "q ∨ p"
using assms
proof
assume "p"
then show "q∨p" by(rule disjI2)
next
assume "q"
show "q∨p" using `q` by(rule disjI1)
qed

text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_28: carmarria
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof -
{assume "p | q"
moreover
{assume p
have "p | r" using `p` by (rule disjI1)
}
moreover
{assume q
have r using assms(1) `q` by (rule mp)
have "p | r" using `r` by (rule disjI2)
}
ultimately have "p | r" by (rule disjE)
}
thus "p | q ⟶ p | r" by (rule impI)
qed

lemma ej_28: joslopjim4
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof (rule impI)
assume "p∨q"
show "p∨r"
using `p∨q`
proof
assume "p"
show "p∨r" using `p` by (rule disjI1)
next
assume "q"
have "r" using assms `q` by (rule mp)
show "p∨r" using `r` by (rule disjI2)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 29. Demostrar
p ∨ p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_29: carmarria
assumes "p ∨ p"
shows "p"
proof -
have "p | p" using assms(1) by this
moreover
{assume p
}
moreover
{assume p
}
ultimately show p by (rule disjE)
qed

lemma ej_29: joslopjim4
assumes "p ∨ p"
shows "p"
proof (rule ccontr)
assume "¬p"
show False
using assms
proof
assume "p"
show False using `¬p` `p` by (rule notE)
next
assume "p"
show False using `¬p` `p` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar
p ⊢ p ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_30: carmarria
assumes "p"
shows "p ∨ p"
proof -
show "p | p" using assms(1) by (rule disjI1)
qed

lemma ej_30: joslopjim4
assumes "p"
shows "p ∨ p"
proof (rule disjI1)
show "p" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_31: carmarria
assumes 1: "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
proof -
have "p | (q | r)" using 1 by this
moreover {assume 2: p
have 3: "p | q" using 2 by (rule disjI1)
have 4: "(p | q) | r" using 3 by (rule disjI1)
}
moreover {assume 5: "q | r"
moreover {assume 6: q
have 7: "p | q" using 6 by (rule disjI2)
have 8: "(p | q) | r " using 7 by (rule disjI1)
}
moreover {assume 9: r
have 10: "(p | q) | r " using 9 by (rule disjI2)
}
ultimately have 11: "(p | q) | r " by (rule disjE)
}
ultimately show "(p | q) | r " by (rule disjE)
qed

lemma ej_31: joslopjim4
assumes "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
using assms
proof
assume "p"
have "p∨q" using `p` by (rule disjI1)
thus "(p ∨ q) ∨ r" by (rule disjI1)
next
assume "q∨r"
show "(p ∨ q) ∨ r"
using `q∨r`
proof
assume "q"
have "p∨q" using `q` by (rule disjI2)
show "(p ∨ q) ∨ r" using `p∨q` by (rule disjI1)
next
assume "r"
show "(p ∨ q) ∨ r" using `r` by (rule disjI2)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_32: carmarria
assumes 1: "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
proof -
have "(p | q) | r" using 1 by this
moreover {assume 2: "p | q"
moreover {assume 3: p
have 4: "p | (q | r)" using 3 by (rule disjI1)
}
moreover {assume 5: q
have 6: "q | r" using 5 by (rule disjI1)
have 7: "p | (q | r)" using 6 by (rule disjI2)
}
ultimately have 8: " p | (q | r)" by (rule disjE)
}
moreover {assume 9: r
have 10: "q | r" using 9 by (rule disjI2)
have 11: "p | (q | r)" using 10 by (rule disjI2)
}
ultimately show "p | (q | r)" by (rule disjE)
qed

lemma ej_32: joslopjim4
assumes "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
using assms
proof
assume "p∨q"
show "p ∨ (q ∨ r)"
using `p∨q`
proof
assume "p"
show "p ∨ (q ∨ r)" using `p` by (rule disjI1)
next
assume "q"
have "q∨r" using `q` by (rule disjI1)
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)
qed
next
assume "r"
have "q∨r" using `r` by (rule disjI2)
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)
qed

text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_33: carmarria
assumes 1: "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 2: p using 1 by (rule conjunct1)
have 3: "q | r" using 1 by (rule conjunct2)
moreover {assume 4: q
have 5: "p & q" using 2 4 by (rule conjI)
have 6: "(p & q) | (p & r)" using 5 by (rule disjI1)
}
moreover {assume 7: r
have 8: "p & r" using 2 7 by (rule conjI)
have 9: "(p & q) | (p & r)" using 8 by (rule disjI2)
}
ultimately show "(p & q) | (p & r)" by (rule disjE)
qed

lemma ej_33: joslopjim4
assumes "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof-
have "p" using assms by (rule conjunct1)
have "q∨r" using assms by (rule conjunct2)
show "(p ∧ q) ∨ (p ∧ r)"
using `q∨r`
proof (rule disjE)
assume "q"
have "p∧q" using `p` `q` by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using `p∧q` by (rule disjI1)
next
assume "r"
have "p∧r" using `p` `r` by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using `p∧r` by (rule disjI2)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_34: carmarria
assumes 1: "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
proof -
have "(p & q) | (p & r)" using 1 by this
moreover {assume 2: "p & q"
have 3: p using 2 by (rule conjunct1)
have 4: q using 2 by (rule conjunct2)
have 5: "q | r" using 4 by (rule disjI1)
have 6: "p & (q | r)" using 3 5 by (rule conjI)
}
moreover {assume 7: "p & r"
have 8: p using 7 by (rule conjunct1)
have 9: r using 7 by (rule conjunct2)
have 10: "q | r" using 9 by (rule disjI2)
have 11: "p & (q | r)" using 8 10 by (rule conjI)
}
ultimately show "p & (q | r)" by (rule disjE)
qed

lemma ej_34: joslopjim4
assumes "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
using assms
proof
assume "p∧q"
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
have "q∨r" using `q` by (rule disjI1)
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)
next
assume "p∧r"
have "p" using `p∧r` by (rule conjunct1)
have "r" using `p∧r` by (rule conjunct2)
have "q∨r" using `r` by (rule disjI2)
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_35: carmarria
assumes 1: "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
proof -
have "p | (q & r)" using 1 by this
moreover {assume 2: p
have 3: " p | q" using 2 by (rule disjI1)
have 4: " p | r" using 2 by (rule disjI1)
have 5: "(p | q) & (p | r)" using 3 4 by (rule conjI)
}
moreover {assume 6: "q & r"
have 7: q using 6 by (rule conjunct1)
have 8: r using 6 by (rule conjunct2)
have 9: "p | q" using 7 by (rule disjI2)
have 10: "p | r" using 8 by (rule disjI2)
have 11: "(p | q) & (p | r)" using 9 10 by (rule conjI)
}
ultimately show "(p | q) & (p | r)" by (rule disjE)
qed

lemma ej_35: joslopjim4
assumes "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
using assms
proof (rule disjE)
assume "p"
have "p∨q" using `p` by (rule disjI1)
have "p∨r" using `p` by (rule disjI1)
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)
next
assume "q∧r"
have "q" using `q∧r` by (rule conjunct1)
have "r" using `q∧r` by (rule conjunct2)
have "p∨q" using `q` by (rule disjI2)
have "p∨r" using `r` by (rule disjI2)
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_36: carmarria
assumes 1: "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof -
have 2: "p | q" using 1 by (rule conjunct1)
moreover {assume 3: p
have 4: "p | (q & r)" using 3 by (rule disjI1)
}
moreover {assume 5: q
have 6: "p | r" using 1 by (rule conjunct2)
moreover {assume 7: p
have 8: "p | (q & r)" using 7 by (rule disjI1)
}
moreover {assume 9: r
have 10: "q & r" using 5 9 by (rule conjI)
have 11: "p | (q & r)" using 10 by (rule disjI2)
}
ultimately have 12: "p | (q & r)" by (rule disjE)
}
ultimately show "p | (q & r)" by (rule disjE)
qed

lemma ej_36: joslopjim4
assumes "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof-
have "p∨q" using assms by (rule conjunct1)
have "p∨r" using assms by (rule conjunct2)
show "p∨(q ∧ r)"
using `p∨r`
proof
assume "p"
show "p∨(q ∧ r)" using `p` by (rule disjI1)
next
assume "r"
show "p∨(q ∧ r)"
using `p∨q`
proof
assume "p"
show "p∨(q ∧ r)" using `p` by (rule disjI1)
next
assume "q"
have "q∧r" using `q` `r` by (rule conjI)
show "p∨(q ∧ r)" using `q∧r` by (rule disjI2)
qed
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_37: carmarria
assumes 1: "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof -
have 2: "p ⟶ r" using 1 by (rule conjunct1)
have 3: "q ⟶ r" using 1 by (rule conjunct2)
{assume 4: "p | q"
moreover {assume 5: p
have 6: r using 2 5 by (rule mp)
}
moreover {assume 7: q
have 8: r using 3 7 by (rule mp)
}
ultimately have r by (rule disjE)
}
thus "p | q ⟶ r" by (rule impI)
qed

lemma ej_37: joslopjim4
assumes "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof (rule impI)
assume "p∨q"
have "p⟶r" using assms by (rule conjunct1)
have "q⟶r" using assms by (rule conjunct2)
show "r"
using `p∨q`
proof
assume "p"
show "r" using `p⟶r` `p` by (rule mp)
next
assume "q"
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 38. Demostrar
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_38: carmarria
assumes 1: "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof -
{assume 2: p
have 3: "p | q" using 2 by (rule disjI1)
have 4: r using 1 3 by (rule mp)
}
hence 5: "p ⟶ r" by (rule impI)
{assume 6: q
have 7: "p | q" using 6 by (rule disjI2)
have 8: r using 1 7 by (rule mp)
}
hence 9: "q ⟶ r" by (rule impI)
show "(p ⟶ r) & (q ⟶ r)" using 5 9 by (rule conjI)
qed

lemma ej_38: joslopjim4
assumes "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof-
have "p⟶r"
proof (rule impI)
assume "p"
have "p∨q" using `p` by (rule disjI1)
show "r" using assms `p∨q` by (rule mp)
qed
have "q⟶r"
proof (rule impI)
assume "q"
have "p∨q" using `q` by (rule disjI2)
show "r" using assms `p∨q` by (rule mp)
qed
show "(p ⟶ r) ∧ (q ⟶ r)" using `p⟶r` `q⟶r` by (rule conjI)
qed

section {* Negaciones *}

text {* ---------------------------------------------------------------
Ejercicio 39. Demostrar
p ⊢ ¬¬p
------------------------------------------------------------------ *}

lemma ejercicio_39: carmarria
assumes "p"
shows "¬¬p"
proof -
show "~~p" using assms(1) by (rule notnotI)
qed

lemma ej_39: joslopjim4
assumes "p"
shows "¬¬p"
proof-
show "¬¬p" using assms by (rule notnotI)
qed

text {* ---------------------------------------------------------------
Ejercicio 40. Demostrar
¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_40: carmarria
assumes "¬p"
shows "p ⟶ q"
proof -
{assume p
have "~p" using assms(1) by this
have False using `~p` `p` by (rule notE)
have q using `False` by (rule FalseE)
}
thus "p ⟶ q" by (rule impI)
qed

lemma ej_40: joslopjim4
assumes "¬p"
shows "p ⟶ q"
proof (rule impI)
assume "p"
show "q" using assms `p` by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 41. Demostrar
p ⟶ q ⊢ ¬q ⟶ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_41: carmarria
assumes 1: "p ⟶ q"
shows "¬q ⟶ ¬p"
proof -
{assume 2: "~q"
have 3: "~p" using 1 2 by (rule mt)
}
thus 4: "~q ⟶ ~p" by (rule impI)
qed

lemma ej_41: joslopjim4
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
proof (rule impI)
assume "¬q"
show "¬p" using assms `¬q` by (rule mt)
qed

text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p∨q, ¬q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_42: carmarria
assumes 1: "p∨q" and
2: "¬q"
shows "p"
proof -
have " p | q" using 1 by this
moreover {assume 3: p
}
moreover {assume 4: q
have 5: False using 2 4 by (rule notE)
have 6: p using 5 by (rule FalseE)
}
ultimately show p by (rule disjE)
qed

lemma ej_42: joslopjim4
assumes "p∨q"
"¬q"
shows "p"
using assms(1)
proof
assume "p"
show "p" using `p` by this
next
assume "q"
show "p" using assms(2) `q` by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 43. Demostrar
p ∨ q, ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_43: carmarria
assumes 1: "p ∨ q" and
2: "¬p"
shows "q"
proof -
have " p | q" using 1 by this
moreover {assume 3: p
have 4: False using 2 3 by (rule notE)
have 5: q using 4 by (rule FalseE)
}
moreover {assume 6: q
}
ultimately show q by (rule disjE)
qed

lemma ej_43: joslopjim4
assumes "p ∨ q"
"¬p"
shows "q"
using assms(1)
proof
assume "p"
show "q" using assms(2) `p` by (rule notE)
next
assume "q"
show "q" using `q` by this
qed

text {* ---------------------------------------------------------------
Ejercicio 44. Demostrar
p ∨ q ⊢ ¬(¬p ∧ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_44: carmarria
assumes 1: "p ∨ q"
shows "¬(¬p ∧ ¬q)"
proof -
have "p | q" using 1 by this
moreover {assume 2: p
{assume 3:"~p & ~q"
have 4: "~p" using 3 by (rule conjunct1)
have 5: False using 4 2 by (rule notE)
}
hence 6: "~(~p & ~q)" by (rule notI)
}
moreover {assume 7: q
{assume 8: "~p & ~q"
have 9: "~q" using 8 by (rule conjunct2)
have 10: False using 9 7 by (rule notE)
}
hence 11: "~(~p & ~q)" by (rule notI)
}
ultimately show "~(~p & ~q)" by (rule disjE)
qed

lemma ej_44: joslopjim4
assumes "p∨q"
shows "¬(¬p∧¬q)"
proof (rule notI)
assume "¬p∧¬q"
show False
using assms
proof
assume "p"
have "¬p" using `¬p∧¬q` by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "q"
have "¬q" using `¬p∧¬q` by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 45. Demostrar
p ∧ q ⊢ ¬(¬p ∨ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_45: carmarria
assumes 1: "p ∧ q"
shows "¬(¬p ∨ ¬q)"
proof -
{assume 2: "~p | ~q"
moreover {assume 3: "~p"
have 4: p using 1 by (rule conjunct1)
have 5: False using 3 4 by (rule notE)
}
moreover {assume 6: "~q"
have 7: q using 1 by (rule conjunct2)
have 8: False using 6 7 by (rule notE)
}
ultimately have False by (rule disjE)
}
thus "~(~p | ~q)" by (rule notI)
qed

lemma ej_45: joslopjim4
assumes "p∧q"
shows "¬(¬p∨¬q)"
proof (rule notI)
assume "¬p∨¬q"
then show False
proof (rule disjE)
assume "¬p"
have "p" using assms by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "¬q"
have "q" using assms by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 46. Demostrar
¬(p ∨ q) ⊢ ¬p ∧ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_46: carmarria
assumes 1: "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof -
{assume 2: p
have 3: "p | q" using 2 by (rule disjI1)
have 4: False using 1 3 by (rule notE)
}
hence 5: "~p" by (rule notI)
{assume 6: q
have 7: "p | q" using 6 by (rule disjI2)
have 8: False using 1 7 by (rule notE)
}
hence 9: "~q" by (rule notI)
show "~p & ~q" using 5 9 by (rule conjI)
qed

lemma ej_46: joslopjim4
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof (rule conjI)
show "¬p"
proof (rule ccontr)
assume "¬¬p"
have "p" using `¬¬p` by (rule notnotD)
have "p∨q" using `p` by (rule disjI1)
show False using assms `p∨q` by (rule notE)
qed
show "¬q"
proof (rule ccontr)
assume "¬¬q"
have "q" using `¬¬q` by (rule notnotD)
have "p∨q" using `q` by (rule disjI2)
show False using assms `p∨q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 47. Demostrar
¬p ∧ ¬q ⊢ ¬(p ∨ q)
------------------------------------------------------------------ *}

lemma ejercicio_47: carmarria
assumes 1: "¬p ∧ ¬q"
shows "¬(p ∨ q)"
proof -
{assume 2: "p | q"
moreover {assume 3: p
have 4: "~p" using 1 by (rule conjunct1)
have 5: False using 4 3 by (rule notE)
}
moreover {assume 6: q
have 7: "~q" using 1 by (rule conjunct2)
have 8: False using 7 6 by (rule notE)
}
ultimately have 9: False by (rule disjE)
}
thus "~(p | q)" by (rule notI)
qed

lemma ej_47: joslopjim4
assumes "¬p∧¬q"
shows "¬(p∨q)"
proof (rule notI)
assume "p∨q"
then show False
proof
assume "p"
have "¬p" using assms by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "q"
have "¬q" using assms by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 48. Demostrar
¬p ∨ ¬q ⊢ ¬(p ∧ q)
------------------------------------------------------------------ *}

lemma ejercicio_48: carmarria
assumes 1: "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof -
have "~p | ~q" using 1 by this
moreover {assume 2: "~p"
{assume 3: "p & q"
have 4: p using 3 by (rule conjunct1)
have 5: False using 2 4 by (rule notE)
}
hence 6: "~(p & q)" by (rule notI)
}
moreover {assume 7: "~q"
{assume 8: "p & q"
have 9: q using 8 by (rule conjunct2)
have 10: False using 7 9 by (rule notE)
}
hence 11: "~(p & q)" by (rule notI)
}
ultimately show "~(p & q)" by (rule disjE)
qed

lemma ej_48: joslopjim4
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof (rule notI)
assume "p∧q"
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
show False
using assms
proof
assume "¬p"
have "p" using `p∧q` by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "¬q"
have "q" using `p∧q` by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 49. Demostrar
⊢ ¬(p ∧ ¬p)
------------------------------------------------------------------ *}

lemma ejercicio_49: carmarria
"¬(p ∧ ¬p)"
proof -
{assume 1: "p & ~p"
have 2: p using 1 by (rule conjunct1)
have 3: "~p" using 1 by (rule conjunct2)
have 4: False using 3 2 by (rule notE)
}
thus "~(p & ~p)" by (rule notI)
qed

lemma ej_49: joslopjim4
shows "¬(p∧¬p)"
proof
assume "p∧¬p"
have "p" using `p∧¬p` by (rule conjunct1)
have "¬p" using `p∧¬p` by (rule conjunct2)
show False using `¬p` `p` by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 50. Demostrar
p ∧ ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_50: carmarria
assumes 1: "p ∧ ¬p"
shows "q"
proof -
have 2: p using 1 by (rule conjunct1)
have 3: "~p" using 1 by (rule conjunct2)
have 4: False using 3 2 by (rule notE)
show q using 4 by (rule FalseE)
qed

lemma ej_50: joslopjim4
assumes "p∧¬p"
shows "q"
proof-
have "p" using `p∧¬p` by (rule conjunct1)
have "¬p" using `p∧¬p` by (rule conjunct2)
have False using `¬p` `p` by (rule notE)
then show "q" by (rule FalseE)
qed

text {* ---------------------------------------------------------------
Ejercicio 51. Demostrar
¬¬p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_51: carmarria joslopjim4
assumes "¬¬p"
shows "p"
proof -
show p using assms by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 52. Demostrar
⊢ p ∨ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_52: carmarria
"p ∨ ¬p"
proof -
{assume 1: "~(p | ~p)"
{assume 2: p
have 3: "p | ~p" using 2 by (rule disjI1)
have 4: False using 1 3 by (rule notE)
}
hence 5: "~p" by (rule notI)
have 6: "p | ~p" using 5 by (rule disjI2)
have 7: False using 1 6 by (rule notE)
}
hence 8: "~~(p | ~p)" by (rule notI)
show "p | ~p" using 8 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 53. Demostrar
⊢ ((p ⟶ q) ⟶ p) ⟶ p
------------------------------------------------------------------ *}

lemma ej_53: joslopjim4
shows "((p ⟶ q) ⟶ p) ⟶ p"
proof (rule impI)
assume "(p ⟶ q) ⟶ p"
show "p"
proof (rule ccontr)
assume "¬p"
have "¬(p⟶q)" using `(p ⟶ q) ⟶ p` `¬p` by (rule mt)
have "p⟶q"
proof (rule impI)
assume "p"
show "q" using `¬p` `p` by (rule notE)
qed
show False using `¬(p⟶q)` `p⟶q` by (rule notE)
qed
qed

lemma ejercicio_53: carmarria
"((p ⟶ q) ⟶ p) ⟶ p"
proof -
{assume 1: "(p ⟶ q) ⟶ p"
{assume 2:"~p"
have 3: "~(p ⟶q)" using 1 2 by (rule mt)
{assume 4: p
have 5: False using 2 4 by (rule notE)
have 6: q using 5 by (rule FalseE)
}
hence 7: "p ⟶q" by (rule impI)
have 8: False using 3 7 by (rule notE)
}
hence 9: "~~p" by (rule notI)
have 10: p using 9 by (rule notnotD)
}
thus "((p ⟶ q)⟶p)⟶p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 54. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ej_54: joslopjim4
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof (rule impI)
assume "p"
have "¬¬p" using `p` by (rule notnotI)
have "¬¬q" using assms `¬¬p` by (rule mt)
show "q" using `¬¬q` by (rule notnotD)
qed

lemma ejercicio_54: carmarria
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{assume 2: p
have 3: "~~p" using 2 by (rule notnotI)
have 4: "~~q" using 1 3 by (rule mt)
have 5: q using 4 by (rule notnotD)
}
thus "p ⟶q" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 55. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_55: carmarria
assumes 1: "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
{assume 2: "~(p | q)"
{assume 3: "~p"
{assume 4: "~q"
have 5: "~p & ~q" using 3 4 by (rule conjI)
have 6: False using 1 5 by (rule notE)
}
hence 7: "~~q" by (rule notI)
have 8: q using 7 by (rule notnotD)
have 9: "p | q" using 8 by (rule disjI2)
have 10: False using 2 9 by (rule notE)
}
hence 11: "~~p" by (rule notI)
have 12: p using 11 by(rule notnotD)
have 13: "p | q" using 12 by (rule disjI1)
have 14: False using 2 13 by (rule notE)
}
hence 15: "~~(p | q)" by (rule notI)
show "p | q" using 15 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 56. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_56: carmarria
assumes 1: "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof -
{assume 2: "~p"
have 3: "~p | ~q" using 2 by (rule disjI1)
have 4: False using 1 3 by (rule notE)
}
hence 5: "~~p" by (rule notI)
have 6: p using 5 by (rule notnotD)
{assume 7: "~q"
have 8: "~p | ~q" using 7 by (rule disjI2)
have 9: False using 1 8 by (rule notE)
}
hence 10: "~~q" by (rule notI)
have 11: q using 10 by (rule notnotD)
show 12: " p & q" using 6 11 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 57. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_57:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
oops

text {* ---------------------------------------------------------------
Ejercicio 58. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_58:
"(p ⟶ q) ∨ (q ⟶ p)"
oops

end

</source>

Relación 4

2018-03-22T14:12:11Z

Carmarria:

<source lang = "isar">

chapter {* R4: Deducción natural proposicional *}

theory R4
imports Main
begin

text {*
---------------------------------------------------------------------
El objetivo de esta relación es demostrar cada uno de los ejercicios
usando sólo las reglas básicas de deducción natural de la lógica
proposicional (sin usar el método auto).

Las reglas básicas de la deducción natural son las siguientes:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· notnotI: P ⟹ ¬¬ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P
---------------------------------------------------------------------
*}

text {*
Se usarán las reglas notnotI y mt que demostramos a continuación. *}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

section {* Implicaciones *}

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
p ⟶ q, p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_1: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ q" and
2: "p"
shows "q"

proof -
show 1: "q" using 1 2 by (rule mp)
qed

lemma ej_1: joslopjim4
assumes "p --> q"
"p"
shows "q"
proof -
show "q" using assms(1) assms(2) by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
p ⟶ q, q ⟶ r, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_2: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ q" and
2: "q ⟶ r" and
3: "p"
shows "r"

proof-
have 4: "q" using 1 3 by (rule mp)
show "r" using 2 4 by (rule mp)
qed

lemma ej_2: joslopjim4
assumes "p ⟶ q"
"q ⟶ r"
"p"
shows "r"
proof -
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_3: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ (q ⟶ r)" and
2: "p ⟶ q" and
3: "p"
shows "r"

proof-
have 4: "q" using 2 3 by (rule mp)
have 5: "(q ⟶ r)" using 1 3 by (rule mp)
show "r" using 5 4 by (rule mp)

qed

lemma ej_3: joslopjim4
assumes "p ⟶ (q ⟶ r)"
"p ⟶ q"
"p"
shows "r"
proof -
have "q" using `p⟶q` `p` by (rule mp)
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
p ⟶ q, q ⟶ r ⊢ p ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_4: josrodjim2
assumes 1: "p ⟶ q" and
2: "q ⟶ r"
shows "p ⟶ r"

proof-
{ assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)}
hence 6: "p ⟶ r" using 3 5 by (rule impI)
show "p⟶r" using 6 by this

oops

lemma ejercicio_4: carmarria
assumes 1: "p ⟶ q" and
2: "q ⟶ r"
shows "p ⟶ r"
proof -
{ assume 3: p
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)
}
thus "p ⟶ r" by (rule impI)
qed
lemma ejercicio_4: inmbenber
assumes 1: "p ⟶ q" and
2: "q ⟶ r"
shows "p ⟶ r"
proof -
{ assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp) }
thus "p ⟶ r" by (rule impI)
qed

lemma ejercicio_4:marmedmar3
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof
assume "p"
with assms(1) have "q" by (rule mp)
with assms(2) show "r" by (rule mp)
qed

lemma ej_4: joslopjim4
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof (rule impI)
assume "p"
have "q" using assms(1) `p` by (rule mp)
show "r" using assms(2) `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_5: josrodjim2 (NO DA ERROR, PERO SI ALERTA)
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

proof -

{ assume 2: "q"
{assume 3: "p"
have 4: "q ⟶r" using 1 3 by (rule mp)
have 5: "r" using 4 2 by (rule mp)}
hence 6: "p⟶r" using 3 5 by (rule impI)}
hence 7: "q ⟶ (p ⟶ r)" using 2 6 by (rule impI)
show "q ⟶ (p ⟶ r)" using 7 by this

oops

lemma ejercicio_5: carmarria
assumes 1: "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

proof -
{assume 2: q
{assume 3: p
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "r" using 4 2 by (rule mp)
}
hence " p ⟶ r" by (rule impI)
}
thus "q ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_5: inmbenber
assumes 1: "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

proof -
{assume 2: "q"
{assume 3: "p"
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "r" using 4 2 by (rule mp) }
hence " p ⟶ r" by (rule impI) }
thus "q ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_5: marmedmar3
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"
proof
assume "q"
show "(p ⟶ r)"
proof
assume "p"
with assms(1) have "(q ⟶ r)" by (rule mp)
thus "r" using `q` by (rule mp)
qed
qed

lemma ej_5: joslopjim4
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"
proof (rule impI)
assume "q"
show "p⟶r"
proof (rule impI)
assume "p"
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 6. Demostrar
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_6:josrodjim2
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"

proof-

{assume 2: "p⟶q"
{assume 3: "p"
have 4: "q⟶r" using 1 3 by (rule mp)
have 5: "q" using 2 3 by (rule mp)
have 6: "r" using 4 5 by (rule mp)}
hence 7: "p⟶r" using 3 6 by (rule impI)}
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" using 2 7 by (rule impI)
show "(p ⟶ q) ⟶ (p ⟶ r)" using 8 by this
oops

lemma ejercicio_6: carmarria, inmbenber
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof -
{assume 2: "p ⟶ q"
{assume 3: "p"
have 4: "q" using 2 3 by (rule mp)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: "r" using 5 4 by (rule mp)
}
hence 7: "p ⟶ r" by (rule impI)
}
thus "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_6:marmedmar3
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof
assume 2: "(p ⟶ q)"
show "(p ⟶ r)"
proof
assume 3: "p"
have 4: "q" using 2 3 by (rule mp)
have 5: "(q ⟶ r)" using 1 3 by (rule mp)
show "r" using 5 4 by (rule mp)
qed
qed

lemma ej_6: joslopjim4
assumes "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof (rule impI)
assume "(p⟶q)"
show "p⟶r"
proof (rule impI)
assume "p"
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
p ⊢ q ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_7: (NO SE QUE ESTA MAL)
assumes "p"
shows "q ⟶ p"

proof -

{ assume 2: "q"
have 3: "p" using 1
}
hence 4: "q⟶p" using 2 3 by (rule impI)

show "q⟶p" using 4 by this

oops

lemma ejercicio_7: carmarria, inmbenber
assumes 1: "p"
shows "q ⟶ p"
proof -
{assume 2: "q"
have 3: "p" using 1 by this}
thus "q ⟶ p" by (rule impI)
qed

lemma ejercicio_7:marmedmar3
assumes "p"
shows "q ⟶ p"
proof
assume "q"
show "p" using assms(1) by this
qed

lemma ej_7: joslopjim4
assumes "p"
shows "q ⟶ p"
proof (rule impI)
assume "q"
show "p" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
⊢ p ⟶ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_8: carmarria
"p ⟶ (q ⟶ p)"

proof -
{assume 1: p
{assume 2: q
have 3: p using 1 by this
}
hence 4: "q ⟶ p" by (rule impI)
}
thus "p ⟶ (q ⟶ p)" by (rule impI)
qed

lemma ejercicio_8: marmedmar3
"p ⟶ (q ⟶ p)"
proof
assume 1: "p"
show "(q ⟶ p)"
proof
assume 2: "q"
show "p" using 1 by this
qed
qed

lemma ej_8: joslopjim4
shows "p⟶(q⟶p)"
proof (rule impI)
assume "p"
show "q⟶p"
proof (rule impI)
assume "q"
show "p" using `p` by this
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_9: carmarria
assumes 1: "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof -
{assume 2: "q ⟶ r"
{assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)
}
hence 6: "p ⟶ r" by (rule impI)
}
thus 7: "(q ⟶ r) ⟶ (p ⟶ r)" by (rule impI)
qed

lemma ejercicio_9: marmedmar3
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof
assume "(q ⟶ r)"
show "(p ⟶ r)"
proof
assume "p"
with `(p ⟶ q)` have "q" by (rule mp)
with `(q ⟶ r)` show "r" by (rule mp)
qed
qed

lemma ej_9: joslopjim4
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof (rule impI)
assume "q⟶r"
show "p⟶r"
proof (rule impI)
assume "p"
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))
------------------------------------------------------------------ *}

lemma ejercicio_10: carmarria
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"

proof-
{assume 2: r
{assume 3: q
{assume 4: p
have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp)
have 6: "r ⟶ s" using 5 3 by (rule mp)
have 7: s using 6 2 by (rule mp)
}
hence 8: "p ⟶ s" by (rule impI)
}
hence 9: "q ⟶ (p ⟶ s)" by (rule impI)
}
thus "r ⟶ (q ⟶ (p ⟶ s))" by (rule impI)
qed

lemma ejercicio_10: marmedmar3
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof
assume 2: "r"
show "(q ⟶ (p ⟶ s))"
proof
assume 3: "q"
show "(p ⟶ s)"
proof
assume 4: "p"
have 5: "(q ⟶ (r ⟶ s))" using 1 4 by (rule mp)
have 6: "(r ⟶ s)" using 5 3 by (rule mp)
show "s" using 6 2 by (rule mp)
qed
qed
qed

lemma ej_10: joslopjim4
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof (rule impI)
assume "r"
show "q⟶(p⟶s)"
proof (rule impI)
assume "q"
show "p⟶s"
proof (rule impI)
assume "p"
have "q ⟶ (r ⟶ s)" using `p ⟶ (q ⟶ (r ⟶ s))` `p` by (rule mp)
have "r⟶s" using `q ⟶ (r ⟶ s)` `q` by (rule mp)
show "s" using `r-->s` `r` by (rule mp)
qed
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))
------------------------------------------------------------------ *}

lemma ejercicio_11: carmarria
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof -
{assume 1: "p ⟶ (q ⟶ r)"
{assume 2: "p ⟶ q"
{assume 3: p
have 4: q using 2 3 by (rule mp)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: r using 5 4 by (rule mp)
}
hence 7: "p ⟶ r" by (rule impI)
}
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)
}
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)
qed

lemma ejercicio_11: marmedmar3
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof
assume 1: "(p ⟶ (q ⟶ r))"
show "((p ⟶ q) ⟶ (p ⟶ r))"
proof
assume 2: "(p ⟶ q)"
show "(p ⟶ r)"
proof
assume 3: "p"
have 4: "(q ⟶ r)" using 1 3 by (rule mp)
have 5: "q" using 2 3 by (rule mp)
show "r" using 4 5 by (rule mp)
qed
qed
qed

lemma ej_11: joslopjim4
shows "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof (rule impI)
assume "p⟶(q⟶r)"
show "(p⟶q)⟶(p⟶r)"
proof (rule impI)
assume "p⟶q"
show "p⟶r"
proof (rule impI)
assume "p"
have "q" using `p⟶q` `p` by (rule mp)
have "q⟶r" using `p⟶(q⟶r)` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_12: carmarria
assumes 1: "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof -
{assume 2: p
{assume 3: q
{assume 4: p
have 5: q using 3 by this
}
hence 6: "p ⟶ q" by (rule impI)
have 7: r using 1 6 by (rule mp)
}
hence 8: "q ⟶ r" by (rule impI)
}
thus "p ⟶ (q ⟶ r)" by (rule impI)
qed

lemma ej_12: joslopjim4
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof (rule impI)
assume "p"
show "q⟶r"
proof (rule impI)
assume "q"
have "p⟶q"
proof (rule impI)
assume "p"
show "q" using `q` by this
qed
show "r" using `(p ⟶ q) ⟶ r` `p⟶q` by (rule mp)
qed
qed

section {* Conjunciones *}

text {* ---------------------------------------------------------------
Ejercicio 13. Demostrar
p, q ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_13: carmarria marmedmar3
assumes "p"
"q"
shows "p ∧ q"

proof-
show "p \<and> q" using assms(1) assms(2) by (rule conjI)
qed

lemma ej_13: joslopjim4
assumes "p"
"q"
shows "p ∧ q"
proof-
show "p∧q" using `p` `q` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
p ∧ q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_14: carmarria marmedmar3
assumes "p ∧ q"
shows "p"

proof -
show "p" using assms by (rule conjunct1)
qed

lemma ej_14: joslopjim4
assumes "p ∧ q"
shows "p"
proof -
show "p" using assms by (rule conjunct1)
qed

text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
p ∧ q ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_15: carmarria marmedmar3
assumes "p ∧ q"
shows "q"

proof -
show q using assms(1) by (rule conjunct2)
qed

lemma ej_15: joslopjim4
assumes "p ∧ q"
shows "q"
proof-
show "q" using assms by (rule conjunct2)
qed

text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_16: carmarria marmedmar3
assumes 1: "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"
proof -
have 2: "q \<and> r" using 1 by (rule conjunct2)
have 3: p using 1 by (rule conjunct1)
have 4: q using 2 by (rule conjunct1)
have 5: r using 2 by (rule conjunct2)
have 6: "p \<and> q" using 3 4 by (rule conjI)
show "(p \<and> q) \<and> r" using 6 5 by (rule conjI)
qed

lemma ej_16: joslopjim4
assumes "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"
proof-
have "p" using assms by (rule conjunct1)
have "q∧r" using assms by (rule conjunct2)
have "q" using `q∧r` by (rule conjunct1)
have "r" using `q∧r` by (rule conjunct2)
have "p∧q" using `p` `q` by (rule conjI)
show "(p∧q)∧r" using `p∧q` `r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 17. Demostrar
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_17:
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof (rule conjI)
have "p ∧ q" using assms ..
thus "p" ..
next
show "q ∧ r"
proof (rule conjI)
have "p ∧ q" using assms ..
thus "q" ..
next
show "r" using assms ..
qed
qed

lemma ejercicio_17: carmarria marmedmar3
assumes 1:"(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof-
have 2:"p \<and> q" using 1 by (rule conjunct1)
have 3: r using 1 by (rule conjunct2)
have 4: p using 2 by (rule conjunct1)
have 5: q using 2 by (rule conjunct2)
have 6: "q \<and> r" using 5 3 by (rule conjI)
show "p \<and> (q \<and> r)" using 4 6 by (rule conjI)
qed

lemma ej_17: joslopjim4
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof-
have "p∧q" using assms by (rule conjunct1)
have "r" using assms by (rule conjunct2)
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
have "q∧r" using `q` `r` by (rule conjI)
show "p∧(q∧r)" using `p` `q∧r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
p ∧ q ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_18: carmarria
assumes "p ∧ q"
shows "p ⟶ q"
proof -
{assume p
have q using assms(1) by (rule conjunct2)
}
thus "p ⟶ q" by (rule impI)
qed

lemma ejercicio_18: marmedmar3
assumes "p ∧ q"
shows "p ⟶ q"
proof
assume "p"
show "q" using assms by (rule conjunct2)
qed

lemma ej_18: joslopjim4
assumes "p ∧ q"
shows "p ⟶ q"
proof (rule impI)
assume "p"
show "q" using assms by (rule conjunct2)
qed

text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_19: carmarria
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof -
{assume 2: p
have 3: "p ⟶ q" using 1 by (rule conjunct1)
have 4: "p ⟶ r" using 1 by (rule conjunct2)
have 5: q using 3 2 by (rule mp)
have 6: r using 4 2 by (rule mp)
have 7: "q \<and> r" using 5 6 by (rule conjI)
}
thus "p ⟶ q \<and> r" by (rule impI)
qed

lemma ejercicio_19: marmedmar3
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof
have 2: "(p ⟶ q)" using assms by (rule conjunct1)
have 3: "(p ⟶ r)" using assms by (rule conjunct2)
assume 4: "p"
with `(p ⟶ q)` have 5: "q" by (rule mp)
have 6: "r" using 3 4 by (rule mp)
show "q ∧ r" using 5 6 by (rule conjI)
qed

lemma ej_19: joslopjim4
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof (rule impI)
assume "p"
have "p⟶q" using assms by (rule conjunct1)
have "q" using `p⟶q` `p` by (rule mp)
have "p⟶r" using assms by (rule conjunct2)
have "r" using `p⟶r` `p` by (rule mp)
show "q∧r" using `q` `r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_20: carmarria
assumes 1: "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof -
{assume 2: p
have 3: "q \<and> r" using 1 2 by (rule mp)
have 4: q using 3 by (rule conjunct1)
}
hence 5: "p ⟶ q" by (rule impI)
{assume 6: p
have 7: "q \<and> r" using 1 6 by (rule mp)
have 8: r using 7 by (rule conjunct2)
}
hence 9: "p ⟶ r" by (rule impI)
show "(p ⟶ q) \<and> (p ⟶ r)" using 5 9 by (rule conjI)
qed

lemma ej_20: joslopjim4
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof-
have "p⟶q"
proof
assume "p"
have "q∧r" using assms `p` by (rule mp)
show "q" using `q∧r` by (rule conjunct1)
qed
have "p⟶r"
proof
assume "p"
have "q∧r" using assms `p` by (rule mp)
show "r" using `q∧r` by (rule conjunct2)
qed
show "(p⟶q)∧(p⟶r)" using `p⟶q` `p⟶r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_21: carmarria
assumes 1: "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof -
{assume 2: "p \<and> q"
have 3: p using 2 by (rule conjunct1)
have 4: q using 2 by (rule conjunct2)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: r using 5 4 by (rule mp)
}
thus "p ∧ q ⟶ r" by (rule impI)
qed

lemma ejercicio_21: marmedmar3
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof
assume "p ∧ q"
hence "p" by (rule conjunct1)
with assms have "(q ⟶ r)" by (rule mp)
have "q" using `p ∧ q` by (rule conjunct2)
with `(q ⟶ r)` show "r" by (rule mp)
qed

lemma ej_21: joslopjim4
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof (rule impI)
assume "p∧q"
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
have "q⟶r" using assms `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_22: carmarria
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof -
{assume p
{assume q
have "p \<and> q" using `p` `q` by (rule conjI)
have r using assms(1) `p \<and> q` by (rule mp)
}
hence "q ⟶ r" by (rule impI)
}
thus "p ⟶ (q ⟶ r)" by (rule impI)
qed

lemma ejercicio_22: marmedmar3
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof
assume "p"
show "(q ⟶ r)"
proof
assume "q"
with `p` have "p ∧ q" by (rule conjI)
with assms show "r" by (rule mp)
qed
qed

lemma ej_22: joslopjim4
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof (rule impI)
assume "p"
show "q⟶r"
proof (rule impI)
assume "q"
have "p∧q" using `p` `q` by (rule conjI)
show "r" using assms `p∧q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_23: carmarria
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof -
{assume "p \<and> q"
{assume p
have q using `p \<and> q` by (rule conjunct2)
}
hence "p ⟶ q" by (rule impI)
have r using assms(1) `p ⟶ q` by (rule mp)
}
thus "p \<and> q ⟶ r" by (rule impI)
qed

lemma ejercicio_23: marmedmar3
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof
assume "p ∧ q"
have "p" using `p ∧ q` by (rule conjunct1)
have "q" using `p ∧ q` by (rule conjunct2)
hence "(p ⟶ q)" by (rule impI)
with assms show "r" by (rule mp)
qed

lemma ej_23: joslopjim4
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof (rule impI)
assume "p∧q"
have "p⟶q"
proof
assume "p"
show "q" using `p∧q` by (rule conjunct2)
qed
show "r" using assms `p⟶q` by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_24: carmarria
assumes 1: "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof -
{assume 2: "p ⟶ q"
have 3: p using 1 by (rule conjunct1)
have 4: "q ⟶ r" using 1 by (rule conjunct2)
have 5: q using 2 3 by (rule mp)
have 6: r using 4 5 by (rule mp)
}
thus "(p ⟶ q) ⟶ r" by (rule impI)
qed

lemma ejercicio_24: marmedmar3
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof
assume "(p ⟶ q)"
have "(q ⟶ r)" using assms by (rule conjunct2)
have "p" using assms by (rule conjunct1)
with `(p ⟶ q)` have "q" by (rule mp)
with `(q ⟶ r)` show "r" by (rule mp)
qed

lemma ej_24: joslopjim4
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof (rule impI)
assume "p⟶q"
have "p" using assms by (rule conjunct1)
have "q⟶r" using assms by (rule conjunct2)
have "q" using `p⟶q` `p` by (rule mp)
show "r" using `q⟶r` `q` by (rule mp)
qed

section {* Disyunciones *}

text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
p ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_25: carmarria
assumes "p"
shows "p ∨ q"
proof -
show "p | q" using assms(1) by (rule disjI1)
qed

lemma ej_25: joslopjim4
assumes "p"
shows "p ∨ q"
proof (rule disjI1)
show "p" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
q ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_26: carmarria
assumes "q"
shows "p ∨ q"
proof -
show "p | q" using assms(1) by (rule disjI2)
qed

lemma ej_26: joslopjim4
assumes "q"
shows "p ∨ q"
proof (rule disjI2)
show "q" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar
p ∨ q ⊢ q ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_27: carmarria
assumes "p ∨ q"
shows "q ∨ p"
proof -
have "p | q" using assms(1) by this
moreover
{assume p
have "q | p" using `p` by (rule disjI2)
}
moreover
{assume q
have "q | p" using `q` by (rule disjI1)
}
ultimately show "q | p" by (rule disjE)
qed

lemma ej_27: joslopjim4
assumes "p ∨ q"
shows "q ∨ p"
using assms
proof
assume "p"
then show "q∨p" by(rule disjI2)
next
assume "q"
show "q∨p" using `q` by(rule disjI1)
qed

text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_28: carmarria
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof -
{assume "p | q"
moreover
{assume p
have "p | r" using `p` by (rule disjI1)
}
moreover
{assume q
have r using assms(1) `q` by (rule mp)
have "p | r" using `r` by (rule disjI2)
}
ultimately have "p | r" by (rule disjE)
}
thus "p | q ⟶ p | r" by (rule impI)
qed

lemma ej_28: joslopjim4
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof (rule impI)
assume "p∨q"
show "p∨r"
using `p∨q`
proof
assume "p"
show "p∨r" using `p` by (rule disjI1)
next
assume "q"
have "r" using assms `q` by (rule mp)
show "p∨r" using `r` by (rule disjI2)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 29. Demostrar
p ∨ p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_29: carmarria
assumes "p ∨ p"
shows "p"
proof -
have "p | p" using assms(1) by this
moreover
{assume p
}
moreover
{assume p
}
ultimately show p by (rule disjE)
qed

lemma ej_29: joslopjim4
assumes "p ∨ p"
shows "p"
proof (rule ccontr)
assume "¬p"
show False
using assms
proof
assume "p"
show False using `¬p` `p` by (rule notE)
next
assume "p"
show False using `¬p` `p` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar
p ⊢ p ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_30: carmarria
assumes "p"
shows "p ∨ p"
proof -
show "p | p" using assms(1) by (rule disjI1)
qed

lemma ej_30: joslopjim4
assumes "p"
shows "p ∨ p"
proof (rule disjI1)
show "p" using assms by this
qed

text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_31: carmarria
assumes 1: "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
proof -
have "p | (q | r)" using 1 by this
moreover {assume 2: p
have 3: "p | q" using 2 by (rule disjI1)
have 4: "(p | q) | r" using 3 by (rule disjI1)
}
moreover {assume 5: "q | r"
moreover {assume 6: q
have 7: "p | q" using 6 by (rule disjI2)
have 8: "(p | q) | r " using 7 by (rule disjI1)
}
moreover {assume 9: r
have 10: "(p | q) | r " using 9 by (rule disjI2)
}
ultimately have 11: "(p | q) | r " by (rule disjE)
}
ultimately show "(p | q) | r " by (rule disjE)
qed

lemma ej_31: joslopjim4
assumes "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
using assms
proof
assume "p"
have "p∨q" using `p` by (rule disjI1)
thus "(p ∨ q) ∨ r" by (rule disjI1)
next
assume "q∨r"
show "(p ∨ q) ∨ r"
using `q∨r`
proof
assume "q"
have "p∨q" using `q` by (rule disjI2)
show "(p ∨ q) ∨ r" using `p∨q` by (rule disjI1)
next
assume "r"
show "(p ∨ q) ∨ r" using `r` by (rule disjI2)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_32: carmarria
assumes 1: "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
proof -
have "(p | q) | r" using 1 by this
moreover {assume 2: "p | q"
moreover {assume 3: p
have 4: "p | (q | r)" using 3 by (rule disjI1)
}
moreover {assume 5: q
have 6: "q | r" using 5 by (rule disjI1)
have 7: "p | (q | r)" using 6 by (rule disjI2)
}
ultimately have 8: " p | (q | r)" by (rule disjE)
}
moreover {assume 9: r
have 10: "q | r" using 9 by (rule disjI2)
have 11: "p | (q | r)" using 10 by (rule disjI2)
}
ultimately show "p | (q | r)" by (rule disjE)
qed

lemma ej_32: joslopjim4
assumes "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
using assms
proof
assume "p∨q"
show "p ∨ (q ∨ r)"
using `p∨q`
proof
assume "p"
show "p ∨ (q ∨ r)" using `p` by (rule disjI1)
next
assume "q"
have "q∨r" using `q` by (rule disjI1)
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)
qed
next
assume "r"
have "q∨r" using `r` by (rule disjI2)
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)
qed

text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_33: carmarria
assumes 1: "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 2: p using 1 by (rule conjunct1)
have 3: "q | r" using 1 by (rule conjunct2)
moreover {assume 4: q
have 5: "p & q" using 2 4 by (rule conjI)
have 6: "(p & q) | (p & r)" using 5 by (rule disjI1)
}
moreover {assume 7: r
have 8: "p & r" using 2 7 by (rule conjI)
have 9: "(p & q) | (p & r)" using 8 by (rule disjI2)
}
ultimately show "(p & q) | (p & r)" by (rule disjE)
qed

lemma ej_33: joslopjim4
assumes "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof-
have "p" using assms by (rule conjunct1)
have "q∨r" using assms by (rule conjunct2)
show "(p ∧ q) ∨ (p ∧ r)"
using `q∨r`
proof (rule disjE)
assume "q"
have "p∧q" using `p` `q` by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using `p∧q` by (rule disjI1)
next
assume "r"
have "p∧r" using `p` `r` by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using `p∧r` by (rule disjI2)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_34: carmarria
assumes 1: "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
proof -
have "(p & q) | (p & r)" using 1 by this
moreover {assume 2: "p & q"
have 3: p using 2 by (rule conjunct1)
have 4: q using 2 by (rule conjunct2)
have 5: "q | r" using 4 by (rule disjI1)
have 6: "p & (q | r)" using 3 5 by (rule conjI)
}
moreover {assume 7: "p & r"
have 8: p using 7 by (rule conjunct1)
have 9: r using 7 by (rule conjunct2)
have 10: "q | r" using 9 by (rule disjI2)
have 11: "p & (q | r)" using 8 10 by (rule conjI)
}
ultimately show "p & (q | r)" by (rule disjE)
qed

lemma ej_34: joslopjim4
assumes "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
using assms
proof
assume "p∧q"
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
have "q∨r" using `q` by (rule disjI1)
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)
next
assume "p∧r"
have "p" using `p∧r` by (rule conjunct1)
have "r" using `p∧r` by (rule conjunct2)
have "q∨r" using `r` by (rule disjI2)
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_35: carmarria
assumes 1: "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
proof -
have "p | (q & r)" using 1 by this
moreover {assume 2: p
have 3: " p | q" using 2 by (rule disjI1)
have 4: " p | r" using 2 by (rule disjI1)
have 5: "(p | q) & (p | r)" using 3 4 by (rule conjI)
}
moreover {assume 6: "q & r"
have 7: q using 6 by (rule conjunct1)
have 8: r using 6 by (rule conjunct2)
have 9: "p | q" using 7 by (rule disjI2)
have 10: "p | r" using 8 by (rule disjI2)
have 11: "(p | q) & (p | r)" using 9 10 by (rule conjI)
}
ultimately show "(p | q) & (p | r)" by (rule disjE)
qed

lemma ej_35: joslopjim4
assumes "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
using assms
proof (rule disjE)
assume "p"
have "p∨q" using `p` by (rule disjI1)
have "p∨r" using `p` by (rule disjI1)
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)
next
assume "q∧r"
have "q" using `q∧r` by (rule conjunct1)
have "r" using `q∧r` by (rule conjunct2)
have "p∨q" using `q` by (rule disjI2)
have "p∨r" using `r` by (rule disjI2)
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_36: carmarria
assumes 1: "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof -
have 2: "p | q" using 1 by (rule conjunct1)
moreover {assume 3: p
have 4: "p | (q & r)" using 3 by (rule disjI1)
}
moreover {assume 5: q
have 6: "p | r" using 1 by (rule conjunct2)
moreover {assume 7: p
have 8: "p | (q & r)" using 7 by (rule disjI1)
}
moreover {assume 9: r
have 10: "q & r" using 5 9 by (rule conjI)
have 11: "p | (q & r)" using 10 by (rule disjI2)
}
ultimately have 12: "p | (q & r)" by (rule disjE)
}
ultimately show "p | (q & r)" by (rule disjE)
qed

lemma ej_36: joslopjim4
assumes "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof-
have "p∨q" using assms by (rule conjunct1)
have "p∨r" using assms by (rule conjunct2)
show "p∨(q ∧ r)"
using `p∨r`
proof
assume "p"
show "p∨(q ∧ r)" using `p` by (rule disjI1)
next
assume "r"
show "p∨(q ∧ r)"
using `p∨q`
proof
assume "p"
show "p∨(q ∧ r)" using `p` by (rule disjI1)
next
assume "q"
have "q∧r" using `q` `r` by (rule conjI)
show "p∨(q ∧ r)" using `q∧r` by (rule disjI2)
qed
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_37: carmarria
assumes 1: "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof -
have 2: "p ⟶ r" using 1 by (rule conjunct1)
have 3: "q ⟶ r" using 1 by (rule conjunct2)
{assume 4: "p | q"
moreover {assume 5: p
have 6: r using 2 5 by (rule mp)
}
moreover {assume 7: q
have 8: r using 3 7 by (rule mp)
}
ultimately have r by (rule disjE)
}
thus "p | q ⟶ r" by (rule impI)
qed

lemma ej_37: joslopjim4
assumes "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof (rule impI)
assume "p∨q"
have "p⟶r" using assms by (rule conjunct1)
have "q⟶r" using assms by (rule conjunct2)
show "r"
using `p∨q`
proof
assume "p"
show "r" using `p⟶r` `p` by (rule mp)
next
assume "q"
show "r" using `q⟶r` `q` by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 38. Demostrar
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_38: carmarria
assumes 1: "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof -
{assume 2: p
have 3: "p | q" using 2 by (rule disjI1)
have 4: r using 1 3 by (rule mp)
}
hence 5: "p ⟶ r" by (rule impI)
{assume 6: q
have 7: "p | q" using 6 by (rule disjI2)
have 8: r using 1 7 by (rule mp)
}
hence 9: "q ⟶ r" by (rule impI)
show "(p ⟶ r) & (q ⟶ r)" using 5 9 by (rule conjI)
qed

lemma ej_38: joslopjim4
assumes "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof-
have "p⟶r"
proof (rule impI)
assume "p"
have "p∨q" using `p` by (rule disjI1)
show "r" using assms `p∨q` by (rule mp)
qed
have "q⟶r"
proof (rule impI)
assume "q"
have "p∨q" using `q` by (rule disjI2)
show "r" using assms `p∨q` by (rule mp)
qed
show "(p ⟶ r) ∧ (q ⟶ r)" using `p⟶r` `q⟶r` by (rule conjI)
qed

section {* Negaciones *}

text {* ---------------------------------------------------------------
Ejercicio 39. Demostrar
p ⊢ ¬¬p
------------------------------------------------------------------ *}

lemma ejercicio_39: carmarria
assumes "p"
shows "¬¬p"
proof -
show "~~p" using assms(1) by (rule notnotI)
qed

lemma ej_39: joslopjim4
assumes "p"
shows "¬¬p"
proof-
show "¬¬p" using assms by (rule notnotI)
qed

text {* ---------------------------------------------------------------
Ejercicio 40. Demostrar
¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_40: carmarria
assumes "¬p"
shows "p ⟶ q"
proof -
{assume p
have "~p" using assms(1) by this
have False using `~p` `p` by (rule notE)
have q using `False` by (rule FalseE)
}
thus "p ⟶ q" by (rule impI)
qed

lemma ej_40: joslopjim4
assumes "¬p"
shows "p ⟶ q"
proof (rule impI)
assume "p"
show "q" using assms `p` by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 41. Demostrar
p ⟶ q ⊢ ¬q ⟶ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_41: carmarria
assumes 1: "p ⟶ q"
shows "¬q ⟶ ¬p"
proof -
{assume 2: "~q"
have 3: "~p" using 1 2 by (rule mt)
}
thus 4: "~q ⟶ ~p" by (rule impI)
qed

lemma ej_41: joslopjim4
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
proof (rule impI)
assume "¬q"
show "¬p" using assms `¬q` by (rule mt)
qed

text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p∨q, ¬q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_42: carmarria
assumes 1: "p∨q" and
2: "¬q"
shows "p"
proof -
have " p | q" using 1 by this
moreover {assume 3: p
}
moreover {assume 4: q
have 5: False using 2 4 by (rule notE)
have 6: p using 5 by (rule FalseE)
}
ultimately show p by (rule disjE)
qed

lemma ej_42: joslopjim4
assumes "p∨q"
"¬q"
shows "p"
using assms(1)
proof
assume "p"
show "p" using `p` by this
next
assume "q"
show "p" using assms(2) `q` by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 43. Demostrar
p ∨ q, ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_43: carmarria
assumes 1: "p ∨ q" and
2: "¬p"
shows "q"
proof -
have " p | q" using 1 by this
moreover {assume 3: p
have 4: False using 2 3 by (rule notE)
have 5: q using 4 by (rule FalseE)
}
moreover {assume 6: q
}
ultimately show q by (rule disjE)
qed

lemma ej_43: joslopjim4
assumes "p ∨ q"
"¬p"
shows "q"
using assms(1)
proof
assume "p"
show "q" using assms(2) `p` by (rule notE)
next
assume "q"
show "q" using `q` by this
qed

text {* ---------------------------------------------------------------
Ejercicio 44. Demostrar
p ∨ q ⊢ ¬(¬p ∧ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_44: carmarria
assumes 1: "p ∨ q"
shows "¬(¬p ∧ ¬q)"
proof -
have "p | q" using 1 by this
moreover {assume 2: p
{assume 3:"~p & ~q"
have 4: "~p" using 3 by (rule conjunct1)
have 5: False using 4 2 by (rule notE)
}
hence 6: "~(~p & ~q)" by (rule notI)
}
moreover {assume 7: q
{assume 8: "~p & ~q"
have 9: "~q" using 8 by (rule conjunct2)
have 10: False using 9 7 by (rule notE)
}
hence 11: "~(~p & ~q)" by (rule notI)
}
ultimately show "~(~p & ~q)" by (rule disjE)
qed

lemma ej_44: joslopjim4
assumes "p∨q"
shows "¬(¬p∧¬q)"
proof (rule notI)
assume "¬p∧¬q"
show False
using assms
proof
assume "p"
have "¬p" using `¬p∧¬q` by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "q"
have "¬q" using `¬p∧¬q` by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 45. Demostrar
p ∧ q ⊢ ¬(¬p ∨ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_45: carmarria
assumes 1: "p ∧ q"
shows "¬(¬p ∨ ¬q)"
proof -
{assume 2: "~p | ~q"
moreover {assume 3: "~p"
have 4: p using 1 by (rule conjunct1)
have 5: False using 3 4 by (rule notE)
}
moreover {assume 6: "~q"
have 7: q using 1 by (rule conjunct2)
have 8: False using 6 7 by (rule notE)
}
ultimately have False by (rule disjE)
}
thus "~(~p | ~q)" by (rule notI)
qed

lemma ej_45: joslopjim4
assumes "p∧q"
shows "¬(¬p∨¬q)"
proof (rule notI)
assume "¬p∨¬q"
then show False
proof (rule disjE)
assume "¬p"
have "p" using assms by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "¬q"
have "q" using assms by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 46. Demostrar
¬(p ∨ q) ⊢ ¬p ∧ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_46: carmarria
assumes 1: "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof -
{assume 2: p
have 3: "p | q" using 2 by (rule disjI1)
have 4: False using 1 3 by (rule notE)
}
hence 5: "~p" by (rule notI)
{assume 6: q
have 7: "p | q" using 6 by (rule disjI2)
have 8: False using 1 7 by (rule notE)
}
hence 9: "~q" by (rule notI)
show "~p & ~q" using 5 9 by (rule conjI)
qed

lemma ej_46: joslopjim4
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof (rule conjI)
show "¬p"
proof (rule ccontr)
assume "¬¬p"
have "p" using `¬¬p` by (rule notnotD)
have "p∨q" using `p` by (rule disjI1)
show False using assms `p∨q` by (rule notE)
qed
show "¬q"
proof (rule ccontr)
assume "¬¬q"
have "q" using `¬¬q` by (rule notnotD)
have "p∨q" using `q` by (rule disjI2)
show False using assms `p∨q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 47. Demostrar
¬p ∧ ¬q ⊢ ¬(p ∨ q)
------------------------------------------------------------------ *}

lemma ejercicio_47: carmarria
assumes 1: "¬p ∧ ¬q"
shows "¬(p ∨ q)"
proof -
{assume 2: "p | q"
moreover {assume 3: p
have 4: "~p" using 1 by (rule conjunct1)
have 5: False using 4 3 by (rule notE)
}
moreover {assume 6: q
have 7: "~q" using 1 by (rule conjunct2)
have 8: False using 7 6 by (rule notE)
}
ultimately have 9: False by (rule disjE)
}
thus "~(p | q)" by (rule notI)
qed

lemma ej_47: joslopjim4
assumes "¬p∧¬q"
shows "¬(p∨q)"
proof (rule notI)
assume "p∨q"
then show False
proof
assume "p"
have "¬p" using assms by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "q"
have "¬q" using assms by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 48. Demostrar
¬p ∨ ¬q ⊢ ¬(p ∧ q)
------------------------------------------------------------------ *}

lemma ejercicio_48: carmarria
assumes 1: "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof -
have "~p | ~q" using 1 by this
moreover {assume 2: "~p"
{assume 3: "p & q"
have 4: p using 3 by (rule conjunct1)
have 5: False using 2 4 by (rule notE)
}
hence 6: "~(p & q)" by (rule notI)
}
moreover {assume 7: "~q"
{assume 8: "p & q"
have 9: q using 8 by (rule conjunct2)
have 10: False using 7 9 by (rule notE)
}
hence 11: "~(p & q)" by (rule notI)
}
ultimately show "~(p & q)" by (rule disjE)
qed

lemma ej_48: joslopjim4
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof (rule notI)
assume "p∧q"
have "p" using `p∧q` by (rule conjunct1)
have "q" using `p∧q` by (rule conjunct2)
show False
using assms
proof
assume "¬p"
have "p" using `p∧q` by (rule conjunct1)
show False using `¬p` `p` by (rule notE)
next
assume "¬q"
have "q" using `p∧q` by (rule conjunct2)
show False using `¬q` `q` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 49. Demostrar
⊢ ¬(p ∧ ¬p)
------------------------------------------------------------------ *}

lemma ejercicio_49: carmarria
"¬(p ∧ ¬p)"
proof -
{assume 1: "p & ~p"
have 2: p using 1 by (rule conjunct1)
have 3: "~p" using 1 by (rule conjunct2)
have 4: False using 3 2 by (rule notE)
}
thus "~(p & ~p)" by (rule notI)
qed

lemma ej_49: joslopjim4
shows "¬(p∧¬p)"
proof
assume "p∧¬p"
have "p" using `p∧¬p` by (rule conjunct1)
have "¬p" using `p∧¬p` by (rule conjunct2)
show False using `¬p` `p` by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 50. Demostrar
p ∧ ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_50: carmarria
assumes 1: "p ∧ ¬p"
shows "q"
proof -
have 2: p using 1 by (rule conjunct1)
have 3: "~p" using 1 by (rule conjunct2)
have 4: False using 3 2 by (rule notE)
show q using 4 by (rule FalseE)
qed

lemma ej_50: joslopjim4
assumes "p∧¬p"
shows "q"
proof-
have "p" using `p∧¬p` by (rule conjunct1)
have "¬p" using `p∧¬p` by (rule conjunct2)
have False using `¬p` `p` by (rule notE)
then show "q" by (rule FalseE)
qed

text {* ---------------------------------------------------------------
Ejercicio 51. Demostrar
¬¬p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_51: carmarria joslopjim4
assumes "¬¬p"
shows "p"
proof -
show p using assms by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 52. Demostrar
⊢ p ∨ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_52: carmarria
"p ∨ ¬p"
proof -
{assume 1: "~(p | ~p)"
{assume 2: p
have 3: "p | ~p" using 2 by (rule disjI1)
have 4: False using 1 3 by (rule notE)
}
hence 5: "~p" by (rule notI)
have 6: "p | ~p" using 5 by (rule disjI2)
have 7: False using 1 6 by (rule notE)
}
hence 8: "~~(p | ~p)" by (rule notI)
show "p | ~p" using 8 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 53. Demostrar
⊢ ((p ⟶ q) ⟶ p) ⟶ p
------------------------------------------------------------------ *}

lemma ej_53: joslopjim4
shows "((p ⟶ q) ⟶ p) ⟶ p"
proof (rule impI)
assume "(p ⟶ q) ⟶ p"
show "p"
proof (rule ccontr)
assume "¬p"
have "¬(p⟶q)" using `(p ⟶ q) ⟶ p` `¬p` by (rule mt)
have "p⟶q"
proof (rule impI)
assume "p"
show "q" using `¬p` `p` by (rule notE)
qed
show False using `¬(p⟶q)` `p⟶q` by (rule notE)
qed
qed

lemma ejercicio_53: carmarria
"((p ⟶ q) ⟶ p) ⟶ p"
proof -
{assume 1: "(p ⟶ q) ⟶ p"
{assume 2:"~p"
have 3: "~(p ⟶q)" using 1 2 by (rule mt)
{assume 4: p
have 5: False using 2 4 by (rule notE)
have 6: q using 5 by (rule FalseE)
}
hence 7: "p ⟶q" by (rule impI)
have 8: False using 3 7 by (rule notE)
}
hence 9: "~~p" by (rule notI)
have 10: p using 9 by (rule notnotD)
}
thus "((p ⟶ q)⟶p)⟶p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 54. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ej_54: joslopjim4
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof (rule impI)
assume "p"
have "¬¬p" using `p` by (rule notnotI)
have "¬¬q" using assms `¬¬p` by (rule mt)
show "q" using `¬¬q` by (rule notnotD)
qed

lemma ejercicio_54: carmarria
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{assume 2: p
have 3: "~~p" using 2 by (rule notnotI)
have 4: "~~q" using 1 3 by (rule mt)
have 5: q using 4 by (rule notnotD)
}
thus "p ⟶q" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 55. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_55: carmarria
assumes 1: "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
{assume 2: "~(p | q)"
{assume 3: "~p"
{assume 4: "~q"
have 5: "~p & ~q" using 3 4 by (rule conjI)
have 6: False using 1 5 by (rule notE)
}
hence 7: "~~q" by (rule notI)
have 8: q using 7 by (rule notnotD)
have 9: "p | q" using 8 by (rule disjI2)
have 10: False using 2 9 by (rule notE)
}
hence 11: "~~p" by (rule notI)
have 12: p using 11 by(rule notnotD)
have 13: "p | q" using 12 by (rule disjI1)
have 14: False using 2 13 by (rule notE)
}
hence 15: "~~(p | q)" by (rule notI)
show "p | q" using 15 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 56. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_56:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
oops

text {* ---------------------------------------------------------------
Ejercicio 57. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_57:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
oops

text {* ---------------------------------------------------------------
Ejercicio 58. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_58:
"(p ⟶ q) ∨ (q ⟶ p)"
oops

end

</source>

Relación 4

2018-03-20T17:02:36Z

Carmarria:

Relación 4

2018-03-20T15:19:13Z

Carmarria:

Relación 4

2018-03-19T15:51:36Z

Carmarria:

Relación 4

2018-03-19T15:50:52Z

Carmarria:

Relación 4

2018-03-16T17:20:52Z

Carmarria:

Relación 4

2018-03-16T16:57:19Z

Carmarria:

Relación 4

2018-03-15T14:33:15Z

Carmarria:

Relación 4

2018-03-15T14:22:43Z

Carmarria:

Relación 4

2018-03-15T14:15:29Z

Carmarria:

Relación 4

2018-03-15T14:09:48Z

Carmarria: