LMF2014: Deducción natural en lógica de primer orden (2)

p>En la clase de hoy del curso Lógica matemática y fundamentos se ha estudiado la segunda parte de la deducción natural en la lógica de primer orden. En concreto, la demostración de equivalencias lógicas y las reglas de la igualdad

p>Las transparencias de estas clases son las páginas 14 a 29 del tema 8.

p>A la vez que se presentaba las reglas, se ha comentado su formalización en Isabelle/HOL. La teoría correspondiente es

header {* Tema 8: Deducción natural en lógica de primer orden *}

theory T8
imports Main 
begin

section {* Demostración de equivalencias *}

text {* 
  Ejemplo 5.1 (p. 15). Demostrar
     ¬∀x. P x  ⊢ ∃x. ¬(P x) *}

-- "La demostración detallada es"
lemma ejemplo_5_1a:
  assumes "¬(∀x. P(x))"
  shows   "∃x. ¬P(x)"
proof (rule ccontr)
  assume "¬(∃x. ¬P(x))"
  have "∀x. P(x)"
  proof (rule allI)
    fix a
    show "P(a)"
    proof (rule ccontr)
      assume "¬P(a)"
      hence "∃x. ¬P(x)" by (rule exI)
      with `¬(∃x. ¬P(x))` show False by (rule notE)
    qed
  qed
  with assms show False by (rule notE)
qed

-- "La demostración estructurada es"
lemma ejemplo_5_1b:
  assumes "¬(∀x. P(x))"
  shows   "∃x. ¬P(x)"
proof (rule ccontr)
  assume "¬(∃x. ¬P(x))"
  have "∀x. P(x)"
  proof 
    fix a
    show "P(a)"
    proof (rule ccontr)
      assume "¬P(a)"
      hence "∃x. ¬P(x)" ..
      with `¬(∃x. ¬P(x))` show False ..
    qed
  qed
  with assms show False ..
qed

-- "La demostración automática es"
lemma ejemplo_5_1c:
  assumes "¬(∀x. P(x))"
  shows   "∃x. ¬P(x)"
using assms
by auto

text {* 
  Ejemplo 5.2 (p. 16). Demostrar
     ∃x. ¬(P x)  ⊢ ¬∀x. P x *}

-- "La demostración detallada es"
lemma ejemplo_5_2a:
  assumes "∃x. ¬P(x)"
  shows   "¬(∀x. P(x))"
proof (rule notI)
  assume "∀x. P(x)"
  obtain a where "¬P(a)" using assms by (rule exE)
  have "P(a)" using `∀x. P(x)` by (rule allE)
  with `¬P(a)` show False by (rule notE)
qed

-- "La demostración estructurada es"
lemma ejemplo_5_2b:
  assumes "∃x. ¬P(x)"
  shows   "¬(∀x. P(x))"
proof 
  assume "∀x. P(x)"
  obtain a where "¬P(a)" using assms ..
  have "P(a)" using `∀x. P(x)` ..
  with `¬P(a)` show False ..
qed

-- "La demostración automática es"
lemma ejemplo_5_2c:
  assumes "∃x. ¬P(x)"
  shows   "¬(∀x. P(x))"
using assms
by auto

text {* 
  Ejemplo 5.3 (p. 17). Demostrar
     ⊢ ¬∀x. P x  ⟷ ∃x. ¬(P x) *}

-- "La demostración detallada es"
lemma ejemplo_5_3a:
  "(¬(∀x. P(x))) ⟷ (∃x. ¬P(x))"
proof (rule iffI)
  assume "¬(∀x. P(x))"
  thus "∃x. ¬P(x)" by (rule ejemplo_5_1a)
next
  assume "∃x. ¬P(x)"
  thus "¬(∀x. P(x))" by (rule ejemplo_5_2a)
qed

-- "La demostración automática es"
lemma ejemplo_5_3b:
  "(¬(∀x. P(x))) ⟷ (∃x. ¬P(x))"
by auto

text {* 
  Ejemplo 6.1 (p. 18). Demostrar
     ∀x. P(x) ∧ Q(x) ⊢  (∀x. P(x)) ∧ (∀x. Q(x)) *}

-- "La demostración detallada es"
lemma ejemplo_6_1a:
  assumes "∀x. P(x) ∧ Q(x)"
  shows   "(∀x. P(x)) ∧ (∀x. Q(x))"
proof (rule conjI)
  show "∀x. P(x)"
  proof (rule allI)
    fix a
    have "P(a) ∧ Q(a)" using assms by (rule allE)
    thus "P(a)" by (rule conjunct1)
  qed
next
  show "∀x. Q(x)"
  proof (rule allI)
    fix a
    have "P(a) ∧ Q(a)" using assms by (rule allE)
    thus "Q(a)" by (rule conjunct2)
  qed
qed

-- "La demostración estructurada es"
lemma ejemplo_6_1b:
  assumes "∀x. P(x) ∧ Q(x)"
  shows   "(∀x. P(x)) ∧ (∀x. Q(x))"
proof 
  show "∀x. P(x)"
  proof 
    fix a
    have "P(a) ∧ Q(a)" using assms ..
    thus "P(a)" ..
  qed
next
  show "∀x. Q(x)"
  proof 
    fix a
    have "P(a) ∧ Q(a)" using assms ..
    thus "Q(a)" ..
  qed
qed

-- "La demostración automática es"
lemma ejemplo_6_1c:
  assumes "∀x. P(x) ∧ Q(x)"
  shows   "(∀x. P(x)) ∧ (∀x. Q(x))"
using assms
by auto

text {* 
  Ejemplo 6.2 (p. 19). Demostrar
     (∀x. P(x)) ∧ (∀x. Q(x)) ⊢ ∀x. P(x) ∧ Q(x)  *}

-- "La demostración detallada es"
lemma ejemplo_6_2a:
  assumes "(∀x. P(x)) ∧ (∀x. Q(x))"
  shows   "∀x. P(x) ∧ Q(x)"
proof (rule allI)
  fix a
  have "∀x. P(x)" using assms by (rule conjunct1)
  hence "P(a)" by (rule allE)
  have "∀x. Q(x)" using assms by (rule conjunct2)
  hence "Q(a)" by (rule allE)
  with `P(a)` show "P(a) ∧ Q(a)" by (rule conjI)
qed

-- "La demostración estructurada es"
lemma ejemplo_6_2b:
  assumes "(∀x. P(x)) ∧ (∀x. Q(x))"
  shows   "∀x. P(x) ∧ Q(x)"
proof
  fix a
  have "∀x. P(x)" using assms ..
  hence "P(a)" by (rule allE)
  have "∀x. Q(x)" using assms ..
  hence "Q(a)" ..
  with `P(a)` show "P(a) ∧ Q(a)" ..
qed

-- "La demostración automática es"
lemma ejemplo_6_2c:
  assumes "(∀x. P(x)) ∧ (∀x. Q(x))"
  shows   "∀x. P(x) ∧ Q(x)"
using assms
by auto

text {* 
  Ejemplo 6.3 (p. 20). Demostrar
     ⊢ ∀x. P(x) ∧ Q(x) ⟷ (∀x. P(x)) ∧ (∀x. Q(x)) *}

-- "La demostración detallada es"
lemma ejemplo_6_3a:
  "(∀x. P(x) ∧ Q(x)) ⟷ ((∀x. P(x)) ∧ (∀x. Q(x)))"
proof (rule iffI)
  assume "∀x. P(x) ∧ Q(x)"
  thus "(∀x. P(x)) ∧ (∀x. Q(x))" by (rule ejemplo_6_1a)
next
  assume "(∀x. P(x)) ∧ (∀x. Q(x))"
  thus "∀x. P(x) ∧ Q(x)" by (rule ejemplo_6_2a)
qed

text {* 
  Ejemplo 7.1 (p. 21). Demostrar
     (∃x. P(x)) ∨ (∃x. Q(x)) ⊢ ∃x. P(x) ∨ Q(x)  *}

-- "La demostración detallada es"
lemma ejemplo_7_1a:
  assumes "(∃x. P(x)) ∨ (∃x. Q(x))"
  shows   "∃x. P(x) ∨ Q(x)"
using assms
proof (rule disjE)
  assume "∃x. P(x)"
  then obtain a where "P(a)" by (rule exE)
  hence "P(a) ∨ Q(a)" by (rule disjI1)
  thus "∃x. P(x) ∨ Q(x)" by (rule exI)
next
  assume "∃x. Q(x)"
  then obtain a where "Q(a)" by (rule exE)
  hence "P(a) ∨ Q(a)" by (rule disjI2)
  thus "∃x. P(x) ∨ Q(x)" by (rule exI)
qed

-- "La demostración estructurada es"
lemma ejemplo_7_1b:
  assumes "(∃x. P(x)) ∨ (∃x. Q(x))"
  shows   "∃x. P(x) ∨ Q(x)"
using assms
proof
  assume "∃x. P(x)"
  then obtain a where "P(a)" ..
  hence "P(a) ∨ Q(a)" ..
  thus "∃x. P(x) ∨ Q(x)" ..
next
  assume "∃x. Q(x)"
  then obtain a where "Q(a)" ..
  hence "P(a) ∨ Q(a)" ..
  thus "∃x. P(x) ∨ Q(x)" ..
qed

-- "La demostración automática es"
lemma ejemplo_7_1c:
  assumes "(∃x. P(x)) ∨ (∃x. Q(x))"
  shows   "∃x. P(x) ∨ Q(x)"
using assms
by auto

text {* 
  Ejemplo 7.2 (p. 22). Demostrar
     ∃x. P(x) ∨ Q(x) ⊢ (∃x. P(x)) ∨ (∃x. Q(x))  *}

-- "La demostración detallada es"
lemma ejemplo_7_2a:
  assumes "∃x. P(x) ∨ Q(x)"
  shows   "(∃x. P(x)) ∨ (∃x. Q(x))"
proof -
  obtain a where "P(a) ∨ Q(a)" using assms by (rule exE)
  thus "(∃x. P(x)) ∨ (∃x. Q(x))"
  proof (rule disjE)
    assume "P(a)"
    hence "∃x. P(x)" by (rule exI)
    thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule disjI1)
  next
    assume "Q(a)"
    hence "∃x. Q(x)" by (rule exI)
    thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule disjI2)
  qed
qed

-- "La demostración estructurada es"
lemma ejercicio_7_2b:
  assumes "∃x. P(x) ∨ Q(x)"
  shows   "(∃x. P(x)) ∨ (∃x. Q(x))"
proof -
  obtain a where "P(a) ∨ Q(a)" using assms ..
  thus "(∃x. P(x)) ∨ (∃x. Q(x))"
  proof 
    assume "P(a)"
    hence "∃x. P(x)" ..
    thus "(∃x. P(x)) ∨ (∃x. Q(x))" ..
  next
    assume "Q(a)"
    hence "∃x. Q(x)" ..
    thus "(∃x. P(x)) ∨ (∃x. Q(x))" ..
  qed
qed

-- "La demostración automática es"
lemma ejercicio_7_2c:
  assumes "∃x. P(x) ∨ Q(x)"
  shows   "(∃x. P(x)) ∨ (∃x. Q(x))"
using assms
by auto

text {* 
  Ejemplo 7.3 (p. 23). Demostrar
     ⊢ ((∃x. P(x)) ∨ (∃x. Q(x))) ⟷ (∃x. P(x) ∨ Q(x))  *}

-- "La demostración detallada es"
lemma ejemplo_7_3a:
  "((∃x. P(x)) ∨ (∃x. Q(x))) ⟷ (∃x. P(x) ∨ Q(x))"
proof (rule iffI)
  assume "(∃x. P(x)) ∨ (∃x. Q(x))"
  thus "∃x. P(x) ∨ Q(x)" by (rule ejemplo_7_1a)
next
  assume "∃x. P(x) ∨ Q(x)"
  thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule ejemplo_7_2a)
qed

-- "La demostración automática es"
lemma ejemplo_7_3b:
  "((∃x. P(x)) ∨ (∃x. Q(x))) ⟷ (∃x. P(x) ∨ Q(x))"
using assms
by auto

text {* 
  Ejemplo 8.1 (p. 24). Demostrar
     ∃x y. P(x,y) ⊢ ∃y x. P(x,y)  *}

-- "La demostración detallada es"
lemma ejemplo_8_1a:
  assumes "∃x y. P(x,y)"
  shows   "∃y x. P(x,y)"
proof -
  obtain a where "∃y. P(a,y)" using assms by (rule exE)
  then obtain b where "P(a,b)" by (rule exE)
  hence "∃x. P(x,b)" by (rule exI)
  thus "∃y x. P(x,y)" by (rule exI)
qed

-- "La demostración estructurada es"
lemma ejemplo_8_1b:
  assumes "∃x y. P(x,y)"
  shows   "∃y x. P(x,y)"
proof -
  obtain a where "∃y. P(a,y)" using assms ..
  then obtain b where "P(a,b)" ..
  hence "∃x. P(x,b)" ..
  thus "∃y x. P(x,y)" ..
qed

-- "La demostración automática es"
lemma ejemplo_8_1c:
  assumes "∃x y. P(x,y)"
  shows   "∃y x. P(x,y)"
using assms
by auto

text {* 
  Ejemplo 8.2. Demostrar
     ∃y x. P(x,y) ⊢ ∃x y. P(x,y)  *}

-- "La demostración detallada es"
lemma ejemplo_8_2a:
  assumes "∃y x. P(x,y)"
  shows   "∃x y. P(x,y)"
proof -
  obtain b where "∃x. P(x,b)" using assms by (rule exE)
  then obtain a where "P(a,b)" by (rule exE)
  hence "∃y. P(a,y)" by (rule exI)
  thus "∃x y. P(x,y)" by (rule exI)
qed

-- "La demostración estructurada es"
lemma ejemplo_8_2b:
  assumes "∃y x. P(x,y)"
  shows   "∃x y. P(x,y)"
proof -
  obtain b where "∃x. P(x,b)" using assms ..
  then obtain a where "P(a,b)" ..
  hence "∃y. P(a,y)" ..
  thus "∃x y. P(x,y)" ..
qed

-- "La demostración estructurada es"
lemma ejemplo_8_2c:
  assumes "∃y x. P(x,y)"
  shows   "∃x y. P(x,y)"
using assms
by auto

text {* 
  Ejemplo 8.3 (p. 25). Demostrar
     ⊢ (∃x y. P(x,y)) ⟷ (∃y x. P(x,y))  *}

-- "La demostración detallada es"
lemma ejemplo_8_3a:
  "(∃x y. P(x,y)) ⟷ (∃y x. P(x,y))"
proof (rule iffI)
  assume "∃x y. P(x,y)"
  thus "∃y x. P(x,y)" by (rule ejemplo_8_1a)
next
  assume "∃y x. P(x,y)"
  thus "∃x y. P(x,y)" by (rule ejemplo_8_2a)
qed

-- "La demostración automática es"
lemma ejemplo_8_3b:
  "(∃x y. P(x,y)) ⟷ (∃y x. P(x,y))"
by auto

section {* Reglas de la igualdad *}

text {*
  Las reglas básicas de la igualdad son:
  · refl:  t = t
  · subst: ⟦s = t; P s⟧ ⟹ P t
*}

text {* 
  Ejemplo 9 (p. 27). Demostrar
     x+1 = 1+x, x+1 > 1 ⟶ x+1 > 0 ⊢ 1+x > 1 ⟶ 1+x > 0
*}

-- "La demostración detallada es"
lemma ejemplo_9a: 
  assumes "x+1 = 1+x" 
          "x+1 > 1 ⟶ x+1 > 0"
  shows   "1+x > 1 ⟶ 1+x > 0"
proof -
  show "1+x > 1 ⟶ 1+x > 0" using assms by (rule subst)
qed

-- "La demostración estructurada es"
lemma ejemplo_9b: 
  assumes "x+1 = 1+x" 
          "x+1 > 1 ⟶ x+1 > 0"
  shows   "1+x > 1 ⟶ 1+x > 0"
using assms 
by (rule subst)

-- "La demostración automática es"
lemma ejemplo_9c: 
  assumes "x+1 = 1+x" 
          "x+1 > 1 ⟶ x+1 > 0"
  shows   "1+x > 1 ⟶ 1+x > 0"
using assms 
by auto

text {* 
  Ejemplo 10 (p. 27). Demostrar
     x = y, y = z ⊢ x = z
*}

-- "La demostración detallada es"
lemma ejemplo_10a:
  assumes "x = y" 
          "y = z"
  shows   "x = z"
proof -
  show "x = z" using assms(2,1) by (rule subst)
qed

-- "La demostración estructurada es"
lemma ejemplo_10b: 
  assumes "x = y" 
          "y = z"
  shows   "x = z"
using assms(2,1)
by (rule subst)

-- "La demostración automática es"
lemma ejemplo_10c: 
  assumes "x = y" 
          "y = z"
  shows   "x = z"
using assms
by auto

text {* 
  Ejemplo 11 (p. 28). Demostrar
     s = t ⊢ t = s
*}

-- "La demostración detallada es"
lemma ejemplo_11a:
  assumes "s = t"
  shows   "t = s"
proof -
  have "s = s" by (rule refl)
  with assms show "t = s" by (rule subst)
qed

-- "La demostración automática es"
lemma ejemplo_11b:
  assumes "s = t"
  shows   "t = s"
using assms
by auto

end

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header {* Tema 8: Deducción natural en lógica de primer orden *}

theory T8

imports Main

begin

section {* Demostración de equivalencias *}

text {*

Ejemplo 5.1 (p. 15). Demostrar

¬∀x. P x ⊢ ∃x. ¬(P x) *}

-- "La demostración detallada es"

lemma ejemplo_5_1a:

assumes "¬(∀x. P(x))"

shows "∃x. ¬P(x)"

proof (rule ccontr)

assume "¬(∃x. ¬P(x))"

have "∀x. P(x)"

proof (rule allI)

fix a

show "P(a)"

proof (rule ccontr)

assume "¬P(a)"

hence "∃x. ¬P(x)" by (rule exI)

with `¬(∃x. ¬P(x))` show False by (rule notE)

qed

with assms show False by (rule notE)

qed

-- "La demostración estructurada es"

lemma ejemplo_5_1b:

assumes "¬(∀x. P(x))"

shows "∃x. ¬P(x)"

proof (rule ccontr)

assume "¬(∃x. ¬P(x))"

have "∀x. P(x)"

proof

fix a

show "P(a)"

proof (rule ccontr)

assume "¬P(a)"

hence "∃x. ¬P(x)" ..

with `¬(∃x. ¬P(x))` show False ..

qed

with assms show False ..

qed

-- "La demostración automática es"

lemma ejemplo_5_1c:

assumes "¬(∀x. P(x))"

shows "∃x. ¬P(x)"

using assms

by auto

text {*

Ejemplo 5.2 (p. 16). Demostrar

∃x. ¬(P x) ⊢ ¬∀x. P x *}

-- "La demostración detallada es"

lemma ejemplo_5_2a:

assumes "∃x. ¬P(x)"

shows "¬(∀x. P(x))"

proof (rule notI)

assume "∀x. P(x)"

obtain a where "¬P(a)" using assms by (rule exE)

have "P(a)" using `∀x. P(x)` by (rule allE)

with `¬P(a)` show False by (rule notE)

qed

-- "La demostración estructurada es"

lemma ejemplo_5_2b:

assumes "∃x. ¬P(x)"

shows "¬(∀x. P(x))"

proof

assume "∀x. P(x)"

obtain a where "¬P(a)" using assms ..

have "P(a)" using `∀x. P(x)` ..

with `¬P(a)` show False ..

qed

-- "La demostración automática es"

lemma ejemplo_5_2c:

assumes "∃x. ¬P(x)"

shows "¬(∀x. P(x))"

using assms

by auto

text {*

Ejemplo 5.3 (p. 17). Demostrar

⊢ ¬∀x. P x ⟷ ∃x. ¬(P x) *}

-- "La demostración detallada es"

lemma ejemplo_5_3a:

"(¬(∀x. P(x))) ⟷ (∃x. ¬P(x))"

proof (rule iffI)

assume "¬(∀x. P(x))"

thus "∃x. ¬P(x)" by (rule ejemplo_5_1a)

assume "∃x. ¬P(x)"

thus "¬(∀x. P(x))" by (rule ejemplo_5_2a)

qed

-- "La demostración automática es"

lemma ejemplo_5_3b:

"(¬(∀x. P(x))) ⟷ (∃x. ¬P(x))"

by auto

text {*

Ejemplo 6.1 (p. 18). Demostrar

∀x. P(x) ∧ Q(x) ⊢ (∀x. P(x)) ∧ (∀x. Q(x)) *}

-- "La demostración detallada es"

lemma ejemplo_6_1a:

assumes "∀x. P(x) ∧ Q(x)"

shows "(∀x. P(x)) ∧ (∀x. Q(x))"

proof (rule conjI)

show "∀x. P(x)"

proof (rule allI)

fix a

have "P(a) ∧ Q(a)" using assms by (rule allE)

thus "P(a)" by (rule conjunct1)

qed

show "∀x. Q(x)"

proof (rule allI)

fix a

have "P(a) ∧ Q(a)" using assms by (rule allE)

thus "Q(a)" by (rule conjunct2)

qed

-- "La demostración estructurada es"

lemma ejemplo_6_1b:

assumes "∀x. P(x) ∧ Q(x)"

shows "(∀x. P(x)) ∧ (∀x. Q(x))"

proof

show "∀x. P(x)"

proof

fix a

have "P(a) ∧ Q(a)" using assms ..

thus "P(a)" ..

qed

show "∀x. Q(x)"

proof

fix a

have "P(a) ∧ Q(a)" using assms ..

thus "Q(a)" ..

qed

-- "La demostración automática es"

lemma ejemplo_6_1c:

assumes "∀x. P(x) ∧ Q(x)"

shows "(∀x. P(x)) ∧ (∀x. Q(x))"

using assms

by auto

text {*

Ejemplo 6.2 (p. 19). Demostrar

(∀x. P(x)) ∧ (∀x. Q(x)) ⊢ ∀x. P(x) ∧ Q(x) *}

-- "La demostración detallada es"

lemma ejemplo_6_2a:

assumes "(∀x. P(x)) ∧ (∀x. Q(x))"

shows "∀x. P(x) ∧ Q(x)"

proof (rule allI)

fix a

have "∀x. P(x)" using assms by (rule conjunct1)

hence "P(a)" by (rule allE)

have "∀x. Q(x)" using assms by (rule conjunct2)

hence "Q(a)" by (rule allE)

with `P(a)` show "P(a) ∧ Q(a)" by (rule conjI)

qed

-- "La demostración estructurada es"

lemma ejemplo_6_2b:

assumes "(∀x. P(x)) ∧ (∀x. Q(x))"

shows "∀x. P(x) ∧ Q(x)"

proof

fix a

have "∀x. P(x)" using assms ..

hence "P(a)" by (rule allE)

have "∀x. Q(x)" using assms ..

hence "Q(a)" ..

with `P(a)` show "P(a) ∧ Q(a)" ..

qed

-- "La demostración automática es"

lemma ejemplo_6_2c:

assumes "(∀x. P(x)) ∧ (∀x. Q(x))"

shows "∀x. P(x) ∧ Q(x)"

using assms

by auto

text {*

Ejemplo 6.3 (p. 20). Demostrar

⊢ ∀x. P(x) ∧ Q(x) ⟷ (∀x. P(x)) ∧ (∀x. Q(x)) *}

-- "La demostración detallada es"

lemma ejemplo_6_3a:

"(∀x. P(x) ∧ Q(x)) ⟷ ((∀x. P(x)) ∧ (∀x. Q(x)))"

proof (rule iffI)

assume "∀x. P(x) ∧ Q(x)"

thus "(∀x. P(x)) ∧ (∀x. Q(x))" by (rule ejemplo_6_1a)

assume "(∀x. P(x)) ∧ (∀x. Q(x))"

thus "∀x. P(x) ∧ Q(x)" by (rule ejemplo_6_2a)

qed

text {*

Ejemplo 7.1 (p. 21). Demostrar

(∃x. P(x)) ∨ (∃x. Q(x)) ⊢ ∃x. P(x) ∨ Q(x) *}

-- "La demostración detallada es"

lemma ejemplo_7_1a:

assumes "(∃x. P(x)) ∨ (∃x. Q(x))"

shows "∃x. P(x) ∨ Q(x)"

using assms

proof (rule disjE)

assume "∃x. P(x)"

then obtain a where "P(a)" by (rule exE)

hence "P(a) ∨ Q(a)" by (rule disjI1)

thus "∃x. P(x) ∨ Q(x)" by (rule exI)

assume "∃x. Q(x)"

then obtain a where "Q(a)" by (rule exE)

hence "P(a) ∨ Q(a)" by (rule disjI2)

thus "∃x. P(x) ∨ Q(x)" by (rule exI)

qed

-- "La demostración estructurada es"

lemma ejemplo_7_1b:

assumes "(∃x. P(x)) ∨ (∃x. Q(x))"

shows "∃x. P(x) ∨ Q(x)"

using assms

proof

assume "∃x. P(x)"

then obtain a where "P(a)" ..

hence "P(a) ∨ Q(a)" ..

thus "∃x. P(x) ∨ Q(x)" ..

assume "∃x. Q(x)"

then obtain a where "Q(a)" ..

hence "P(a) ∨ Q(a)" ..

thus "∃x. P(x) ∨ Q(x)" ..

qed

-- "La demostración automática es"

lemma ejemplo_7_1c:

assumes "(∃x. P(x)) ∨ (∃x. Q(x))"

shows "∃x. P(x) ∨ Q(x)"

using assms

by auto

text {*

Ejemplo 7.2 (p. 22). Demostrar

∃x. P(x) ∨ Q(x) ⊢ (∃x. P(x)) ∨ (∃x. Q(x)) *}

-- "La demostración detallada es"

lemma ejemplo_7_2a:

assumes "∃x. P(x) ∨ Q(x)"

shows "(∃x. P(x)) ∨ (∃x. Q(x))"

proof -

obtain a where "P(a) ∨ Q(a)" using assms by (rule exE)

thus "(∃x. P(x)) ∨ (∃x. Q(x))"

proof (rule disjE)

assume "P(a)"

hence "∃x. P(x)" by (rule exI)

thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule disjI1)

assume "Q(a)"

hence "∃x. Q(x)" by (rule exI)

thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule disjI2)

qed

-- "La demostración estructurada es"

lemma ejercicio_7_2b:

assumes "∃x. P(x) ∨ Q(x)"

shows "(∃x. P(x)) ∨ (∃x. Q(x))"

proof -

obtain a where "P(a) ∨ Q(a)" using assms ..

thus "(∃x. P(x)) ∨ (∃x. Q(x))"

proof

assume "P(a)"

hence "∃x. P(x)" ..

thus "(∃x. P(x)) ∨ (∃x. Q(x))" ..

assume "Q(a)"

hence "∃x. Q(x)" ..

thus "(∃x. P(x)) ∨ (∃x. Q(x))" ..

qed

-- "La demostración automática es"

lemma ejercicio_7_2c:

assumes "∃x. P(x) ∨ Q(x)"

shows "(∃x. P(x)) ∨ (∃x. Q(x))"

using assms

by auto

text {*

Ejemplo 7.3 (p. 23). Demostrar

⊢ ((∃x. P(x)) ∨ (∃x. Q(x))) ⟷ (∃x. P(x) ∨ Q(x)) *}

-- "La demostración detallada es"

lemma ejemplo_7_3a:

"((∃x. P(x)) ∨ (∃x. Q(x))) ⟷ (∃x. P(x) ∨ Q(x))"

proof (rule iffI)

assume "(∃x. P(x)) ∨ (∃x. Q(x))"

thus "∃x. P(x) ∨ Q(x)" by (rule ejemplo_7_1a)

assume "∃x. P(x) ∨ Q(x)"

thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule ejemplo_7_2a)

qed

-- "La demostración automática es"

lemma ejemplo_7_3b:

"((∃x. P(x)) ∨ (∃x. Q(x))) ⟷ (∃x. P(x) ∨ Q(x))"

using assms

by auto

text {*

Ejemplo 8.1 (p. 24). Demostrar

∃x y. P(x,y) ⊢ ∃y x. P(x,y) *}

-- "La demostración detallada es"

lemma ejemplo_8_1a:

assumes "∃x y. P(x,y)"

shows "∃y x. P(x,y)"

proof -

obtain a where "∃y. P(a,y)" using assms by (rule exE)

then obtain b where "P(a,b)" by (rule exE)

hence "∃x. P(x,b)" by (rule exI)

thus "∃y x. P(x,y)" by (rule exI)

qed

-- "La demostración estructurada es"

lemma ejemplo_8_1b:

assumes "∃x y. P(x,y)"

shows "∃y x. P(x,y)"

proof -

obtain a where "∃y. P(a,y)" using assms ..

then obtain b where "P(a,b)" ..

hence "∃x. P(x,b)" ..

thus "∃y x. P(x,y)" ..

qed

-- "La demostración automática es"

lemma ejemplo_8_1c:

assumes "∃x y. P(x,y)"

shows "∃y x. P(x,y)"

using assms

by auto

text {*

Ejemplo 8.2. Demostrar

∃y x. P(x,y) ⊢ ∃x y. P(x,y) *}

-- "La demostración detallada es"

lemma ejemplo_8_2a:

assumes "∃y x. P(x,y)"

shows "∃x y. P(x,y)"

proof -

obtain b where "∃x. P(x,b)" using assms by (rule exE)

then obtain a where "P(a,b)" by (rule exE)

hence "∃y. P(a,y)" by (rule exI)

thus "∃x y. P(x,y)" by (rule exI)

qed

-- "La demostración estructurada es"

lemma ejemplo_8_2b:

assumes "∃y x. P(x,y)"

shows "∃x y. P(x,y)"

proof -

obtain b where "∃x. P(x,b)" using assms ..

then obtain a where "P(a,b)" ..

hence "∃y. P(a,y)" ..

thus "∃x y. P(x,y)" ..

qed

-- "La demostración estructurada es"

lemma ejemplo_8_2c:

assumes "∃y x. P(x,y)"

shows "∃x y. P(x,y)"

using assms

by auto

text {*

Ejemplo 8.3 (p. 25). Demostrar

⊢ (∃x y. P(x,y)) ⟷ (∃y x. P(x,y)) *}

-- "La demostración detallada es"

lemma ejemplo_8_3a:

"(∃x y. P(x,y)) ⟷ (∃y x. P(x,y))"

proof (rule iffI)

assume "∃x y. P(x,y)"

thus "∃y x. P(x,y)" by (rule ejemplo_8_1a)

assume "∃y x. P(x,y)"

thus "∃x y. P(x,y)" by (rule ejemplo_8_2a)

qed

-- "La demostración automática es"

lemma ejemplo_8_3b:

"(∃x y. P(x,y)) ⟷ (∃y x. P(x,y))"

by auto

section {* Reglas de la igualdad *}

text {*

Las reglas básicas de la igualdad son:

· refl: t = t

· subst: ⟦s = t; P s⟧ ⟹ P t

text {*

Ejemplo 9 (p. 27). Demostrar

x+1 = 1+x, x+1 > 1 ⟶ x+1 > 0 ⊢ 1+x > 1 ⟶ 1+x > 0

-- "La demostración detallada es"

lemma ejemplo_9a:

assumes "x+1 = 1+x"

"x+1 > 1 ⟶ x+1 > 0"

shows "1+x > 1 ⟶ 1+x > 0"

proof -

show "1+x > 1 ⟶ 1+x > 0" using assms by (rule subst)

qed

-- "La demostración estructurada es"

lemma ejemplo_9b:

assumes "x+1 = 1+x"

"x+1 > 1 ⟶ x+1 > 0"

shows "1+x > 1 ⟶ 1+x > 0"

using assms

by (rule subst)

-- "La demostración automática es"

lemma ejemplo_9c:

assumes "x+1 = 1+x"

"x+1 > 1 ⟶ x+1 > 0"

shows "1+x > 1 ⟶ 1+x > 0"

using assms

by auto

text {*

Ejemplo 10 (p. 27). Demostrar

x = y, y = z ⊢ x = z

-- "La demostración detallada es"

lemma ejemplo_10a:

assumes "x = y"

"y = z"

shows "x = z"

proof -

show "x = z" using assms(2,1) by (rule subst)

qed

-- "La demostración estructurada es"

lemma ejemplo_10b:

assumes "x = y"

"y = z"

shows "x = z"

using assms(2,1)

by (rule subst)

-- "La demostración automática es"

lemma ejemplo_10c:

assumes "x = y"

"y = z"

shows "x = z"

using assms

by auto

text {*

Ejemplo 11 (p. 28). Demostrar

s = t ⊢ t = s

-- "La demostración detallada es"

lemma ejemplo_11a:

assumes "s = t"

shows "t = s"

proof -

have "s = s" by (rule refl)

with assms show "t = s" by (rule subst)

qed

-- "La demostración automática es"

lemma ejemplo_11b:

assumes "s = t"

shows "t = s"

using assms

by auto

end

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