Answer to Question #233865 in Discrete Mathematics for bujji babu

Question #233865

Show that ¬ (P    \iffQ)    \iff(P V Q) Λ ¬(P Λ Q)     \iff(P Λ ¬Q) V (¬ P Λ Q) without using truth table


1
Expert's answer
2021-10-18T13:41:18-0400

Solution:

¬(PQ)¬((PQ)(QP))¬((P¬Q)(Q¬P))¬(((P¬Q)Q)((P¬Q)¬P))¬((PQ)(¬QQ)((P¬P)(¬Q¬P))¬((PQ)¬(PQ))¬(PQ)¬(¬(PQ))(PQ)¬(PQ)(1)\begin{aligned} &\neg(P \Leftrightarrow Q) \\ &\Leftrightarrow \neg((P \rightarrow Q) \wedge(Q \rightarrow P)) \\ &\Leftrightarrow \neg((P \vee \neg Q) \wedge(Q \vee \neg P)) \\ &\Leftrightarrow \neg(((P \vee \neg Q) \wedge Q) \vee((P \vee \neg Q) \wedge \neg P)) \\ &\Leftrightarrow \neg((P \wedge Q) \vee(\neg Q \wedge Q) \vee((P \wedge \neg P) \vee(\neg Q \wedge \neg P)) \\ &\Leftrightarrow \neg((P \wedge Q) \vee \neg(P \vee Q)) \\ &\Leftrightarrow \neg(P \wedge Q) \wedge \neg(\neg(P \vee Q)) \\ &\Leftrightarrow(P \vee Q) \wedge \neg(P \wedge Q) \ldots(1) \end{aligned}

Also,

¬(PQ)¬((PQ)(QP))¬((P¬Q)(Q¬P))(¬PQ)(¬QP)(2)\begin{aligned} &\neg(P \Leftrightarrow Q) \\ &\Leftrightarrow \neg((P \rightarrow Q) \wedge(Q \rightarrow P)) \\ &\Leftrightarrow \neg((P \vee \neg Q) \wedge(Q \vee \neg P)) \\ &\Leftrightarrow(\neg P \wedge Q) \vee(\neg Q \wedge P) \ldots(2) \end{aligned}

Therefore the result follows from (1) and (2).


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