KB = (B ↔ p v q) & ~ B α= ~ p
(B ↔ p v q) & ~ B
KB = (B ↔ p v q) & ~ B α= ~ p (B ↔ p v q) & ~ B
KB = (B ↔ p v q) & ~ B α= ~ p (B ↔ p v q) & ~ B
KB = (B ↔ p v q) & ~ B α= ~ p (B ↔ p v q) & ~ B ( (B →(p v q)) & ( (p v q) → B ) )& ~ B
KB = (B ↔ p v q) & ~ B α= ~ 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) & ~ B α= ~ 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) & ~ B α= ~ 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) & ~ B α= ~ 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) & ~ B α= ~ p ~B v p v q ~p v B ~q v B ~B
KB = (B ↔ p v q) & ~ B α= ~ p ~B v p v q ~p v B ~q v B ~B p
KB = (B ↔ p v q) & ~ B α= ~ p ~B v p v q ~p v B ~q v B ~B p ~p
KB = (B ↔ p v q) & ~ B α= ~ p ~B v p v q ~p v B ~q v B ~B p ~p