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KB = (B ↔ p v q) & ~ B α = ~ p. 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

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kb b p v q b p4
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 b p5
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 b p6
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 b p7
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 b p8
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 b p9
KB = (B ↔ p v q) & ~ Bα= ~ p

~B

~q v B

~B v p v q

~p v B

kb b p v q b p10
KB = (B ↔ p v q) & ~ Bα= ~ p

~B

p

~q v B

~B v p v q

~p v B

kb b p v q b p11
KB = (B ↔ p v q) & ~ Bα= ~ p

~B

p

~q v B

~B v p v q

~p v B

~p

kb b p v q b p12
KB = (B ↔ p v q) & ~ Bα= ~ p

~B

p

~q v B

~B v p v q

~p v B

~p