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Can someone explain how in propositional logic these are equivalent :

A ∧ B ∧ (¬B ∨ ¬C) ≡ A ∧ B ∧ ¬C

Because using the distributive law I would get:

 ≡ (A ∧ B ∧ ¬B) ∨ (A ∧ B ∧ ¬C)
 ≡  A ∨ (A ∧ B ∧ ¬C)
 ≡ (A ∨ A) ∧ (A ∨ B) ∧ (A ∨ ¬C)
 ≡  A ∧ (A ∨ B) ∧ (A ∨ ¬C)
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Hint. B ∧ ¬B is false, so what is A ∧ B ∧ ¬B ?

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  • $\begingroup$ Oh I see, so A ∧ B ∧ ¬B must be false , leaving just A ∧ B ∧ ¬C $\endgroup$ – xava May 23 at 9:16
  • $\begingroup$ Well done, you got it. $\endgroup$ – J.-E. Pin May 23 at 9:24
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Your first use of the distributive law was correct. Now note that the statement $A\wedge B\wedge \neg B$ is a contradiction, and therefore we have $$\begin{array}{rcl}A\wedge B \wedge (\neg B \vee \neg C) &\equiv& (A\wedge B \wedge \neg B) \vee (A \wedge B \wedge \neg C)\\&\equiv &(A\wedge (B \wedge \neg B)) \vee (A \wedge B \wedge \neg C)\\ &\equiv& (A\wedge F)\vee (A \wedge B \wedge \neg C)\\& \equiv&F \vee (A \wedge B \wedge \neg C)\\&\equiv& (A \wedge B \wedge \neg C) \end{array}$$ Where $F$ denotes falsem.

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