KB = (B ↔ p v q) & ~ B α = ~ p
description
Transcript of KB = (B ↔ p v q) & ~ B α = ~ p
KB = (B ↔ p v q) & ~ Bα= ~ p
KB = (B ↔ p v q) & ~ Bα= ~ p
(B ↔ p v q) & ~ B
KB = (B ↔ p v q) & ~ Bα= ~ p
(B ↔ p v q) & ~ B
KB = (B ↔ p v q) & ~ Bα= ~ p
(B ↔ p v q) & ~ B
KB = (B ↔ p v q) & ~ Bα= ~ p
(B ↔ p v q) & ~ B
( (B →(p v q)) & ((p v q) →B) ) & ~ B
KB = (B ↔ p v q) & ~ Bα= ~ p
(B ↔ p v q) & ~ B
( (B →(p v q)) & ((p v q) →B) ) & ~ B
( (~B v(p v q)) & (~(p v q) v B) ) & ~ B
KB = (B ↔ p v q) & ~ Bα= ~ p
(B ↔ p v q) & ~ B
( (B →(p v q)) & ((p v q) →B) ) & ~ B
( (~B v p v q) & ((~p & ~q) v B) ) & ~ B
KB = (B ↔ p v q) & ~ Bα= ~ p
(B ↔ p v q) & ~ B
( (B →(p v q)) & ((p v q) →B) ) & ~ B
( (~B v p v q) & ((~p & ~q) v B) ) & ~ B
(~B v p v q) & (~p v B) & (~q v B) & ~ B
KB = (B ↔ p v q) & ~ Bα= ~ p
(B ↔ p v q) & ~ B
( (B →(p v q)) & ((p v q) →B) ) & ~ B
( (~B v p v q) & ((~p & ~q) v B) ) & ~ B
(~B v p v q) & (~p v B) & (~q v B) & ~ B
KB = (B ↔ p v q) & ~ Bα= ~ p
~B v p v q ~p v B ~q v B ~B
KB = (B ↔ p v q) & ~ Bα= ~ p
~B v p v q ~p v B ~q v B ~B p
KB = (B ↔ p v q) & ~ Bα= ~ p
~B v p v q ~p v B ~q v B ~B p
~p
KB = (B ↔ p v q) & ~ Bα= ~ p
~B v p v q ~p v B ~q v B ~B p
~p