Answer to Question #218792 in Discrete Mathematics for ratnakar

Question

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

Answer 1

Since

$\neg \left( {P \leftrightarrow Q} \right) \equiv \neg \left( {\left( {P \to Q} \right) \wedge \left( {Q \to P} \right)} \right) \equiv \neg \left( {\left( {\neg P \vee Q} \right) \wedge \left( {\neg Q \vee P} \right)} \right) \equiv \neg \left( {\neg P \vee Q} \right) \vee \neg \left( {\neg Q \vee P} \right) \equiv \left( {P \wedge \neg Q} \right) \vee \left( {Q \wedge \neg P} \right)$

Then

$\neg \left( {P \leftrightarrow Q} \right) \leftrightarrow \left( {P \wedge \neg Q} \right) \vee \left( {\neg P \wedge Q} \right)$

Since

$\left( {P \wedge \neg Q} \right) \vee \left( {\neg P \wedge Q} \right) \equiv \left( {P \vee \neg P} \right) \wedge \left( {\neg Q \vee \neg P} \right) \wedge \left( {P \vee Q} \right) \wedge \left( {\neg Q \vee Q} \right) \equiv 1 \wedge \left( {\neg Q \vee \neg P} \right) \wedge \left( {P \vee Q} \right) \wedge 1 \equiv \left( {\neg Q \vee \neg P} \right) \wedge \left( {P \vee Q} \right) \equiv \left( {P \vee Q} \right) \wedge \neg \left( {P \wedge Q} \right)$

Then

$\left( {P \wedge \neg Q} \right) \vee \left( {\neg P \wedge Q} \right) \leftrightarrow \left( {P \vee Q} \right) \wedge \neg \left( {P \wedge Q} \right)$

But then

$\neg \left( {P \leftrightarrow Q} \right) \leftrightarrow \left( {P \vee Q} \right) \wedge \neg \left( {P \wedge Q} \right) \leftrightarrow \left( {P \wedge \neg Q} \right) \vee \left( {\neg P \wedge Q} \right)$

Q. E. D.