Diferencia entre revisiones de «Relación 1»
De Razonamiento automático (2010-11)
Línea 36: | Línea 36: | ||
lemma "(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))" | lemma "(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))" | ||
− | + | proof | |
+ | assume 1:"(A ⟶ B)" | ||
+ | show "((B ⟶ C) ⟶ (A ⟶ C))" | ||
+ | proof | ||
+ | assume 2:"(B ⟶ C)" | ||
+ | show "(A ⟶ C)" | ||
+ | proof | ||
+ | assume 3:"A" | ||
+ | from 1 3 have 4:"B" by (rule mp) | ||
+ | from 2 4 show "C" by (rule mp) | ||
+ | qed | ||
+ | qed | ||
+ | qed | ||
lemma "¬¬A ⟶ A" | lemma "¬¬A ⟶ A" |
Revisión del 18:02 1 feb 2011
header {* 1ª relación de ejercicios *}
theory Relacion_1
imports Main
begin
text {*
---------------------------------------------------------------------
Demostrar los siguientes lemas usando sólo las reglas básicas de
deducción natural de la lógica proposicional.
---------------------------------------------------------------------
*}
lemma I: "A ⟶ A"
oops
lemma "A ∧ B ⟶ B ∧ A"
oops
lemma "A ∧ B ⟶ A ∨ B"
proof
assume "A ∧ B"
hence "A" ..
thus "A ∨ B" ..
qed
lemma "(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)"
oops
lemma K: "A ⟶ (B ⟶ A)"
oops
lemma S: "(A ⟶ (B ⟶ C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))"
oops
lemma "(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))"
proof
assume 1:"(A ⟶ B)"
show "((B ⟶ C) ⟶ (A ⟶ C))"
proof
assume 2:"(B ⟶ C)"
show "(A ⟶ C)"
proof
assume 3:"A"
from 1 3 have 4:"B" by (rule mp)
from 2 4 show "C" by (rule mp)
qed
qed
qed
lemma "¬¬A ⟶ A"
oops
lemma "A ⟶ ¬¬A"
oops
lemma "(¬A ⟶ B) ⟶ (¬B ⟶ A)"
oops
lemma "((A ⟶ B) ⟶ A) ⟶ A"
oops
lemma "A ∨ ¬A"
oops
lemma "(¬(A ∧ B)) = (¬A ∨ ¬B)"
oops
end