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\( p \leftrightarrow q \) is logically NOT equivalent to (A) \( (\sim p \vee q ) \wedge(\sim q \vee p ) \) (B) \( (p \wedge q) \vee(\sim p \wedge \sim q) \) (C) \( (p \wedge \sim q) \vee(q \wedge \sim p) \) (D) \( (p \rightarrow q) \wedge(q \rightarrow p) \) |
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Answer» (i) p ↔ q = p ↔ q and q ↔ p = (p ↔ q) ^ (q ↔ p) Hence, p ↔ q = (p ↔ q) ^ (q ↔ p) It means p ↔ q is logically equivalent to (p ↔ q) ^ (q ↔ p) (ii) p ↔ q = q ↔ p and p ↔ q = ~ p ↔ ~q and ~q ↔ ~p = ~(~p) V ~ q and ~(~q) V ~p = (p v ~q) ^ (q v ~p) = (~pvq) ^ (~qvp) Hence, (A) is also equivalent to p ↔ q (iii) p ↔ q = p ↔ q and q ↔ p = (~pvq) ^ (~qvp) = ((~pvq) ^ ~q) V ((~pvq) ^ p) = ((~ p ^ ~q)) v (q ^ ~q)) v ((~P^P)) V (q^p)) = ((~p ^ ~q)) V F) V (FV(P ^q)) (∵ ~p ^ P = F) = (~p ^ ~q) V (P ^ q) Hence, (B) is equivalent to p ↔ q. i.e, (p ^ ~q) V (q ^ ~p) is not logically equivalent to p ↔ q. |
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