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KB = (B ↔ p v q) & ~ B α = ~ p

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 α = ~ p

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  1. KB = (B ↔ p v q) & ~ Bα= ~ p

  2. KB = (B ↔ p v q) & ~ Bα= ~ p (B ↔ p v q) & ~ B

  3. KB = (B ↔ p v q) & ~ Bα= ~ p (B ↔ p v q) & ~ B

  4. KB = (B ↔ p v q) & ~ Bα= ~ p (B ↔ p v q) & ~ B

  5. KB = (B ↔ p v q) & ~ Bα= ~ p (B ↔ p v q) & ~ B ( (B →(p v q)) &((p v q) →B) ) & ~ B

  6. 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

  7. 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

  8. 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

  9. 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

  10. KB = (B ↔ p v q) & ~ Bα= ~ p ~B ~q v B ~B v p v q ~p v B

  11. KB = (B ↔ p v q) & ~ Bα= ~ p ~B p ~q v B ~B v p v q ~p v B

  12. KB = (B ↔ p v q) & ~ Bα= ~ p ~B p ~q v B ~B v p v q ~p v B ~p

  13. KB = (B ↔ p v q) & ~ Bα= ~ p ~B p ~q v B ~B v p v q ~p v B ~p

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