Relación 1
De Razonamiento automático (2010-11)
Revisión del 18:36 1 feb 2011 de Palmagroblanco (discusión | contribuciones)
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"
by (rule imp_refl)
lemma
assumes 1:"A ∧ B"
shows "B ∧ A"
proof -
from 1 have 2:"A" by (rule conjunct1)
from 1 have 3:"B" by (rule conjunct2)
from 3 2 show "B ∧ A" by (rule conjI)
qed
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