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The Conjunctive Normal form of a formula $F$  is

$(P \vee Q \vee P)  \wedge  (P \vee Q \vee Q)   \wedge   (¬P \vee ¬Q \vee ¬P)   \wedge  (¬P \vee ¬Q \vee ¬Q)$. where, $\wedge$ means AND, $\vee$ means OR

Then the value of $F$  is:

  1. $T$
  2. $¬(P  \wedge Q) \leftrightarrow (P \vee Q)$
  3. $(P \vee Q) \leftrightarrow (P \wedge Q)$
  4. $¬(P  \wedge Q) \to (P \vee Q )$
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$(P \vee Q \vee P)  \wedge  (P \vee Q \vee Q)   \wedge   (¬P \vee ¬Q \vee ¬P)   \wedge  (¬P \vee ¬Q \vee ¬Q)$

$=(P \vee Q )  \wedge (¬P \vee ¬Q)$

$=(¬P \to Q)\wedge  (P \to ¬Q)$

$=(¬Q \to P)\wedge (P \to ¬Q)$

$=P\leftrightarrow \neg Q$
 

Now if we reduce option B, we get

$\neg(P \wedge Q) \leftrightarrow (P \vee Q)$

$= ((P \wedge Q) \vee (P \vee Q)) \wedge ( \neg (P \vee Q) \vee \neg(P \wedge Q))$

$= (P \vee Q) \wedge \neg (P \wedge Q)$

$ = (P \vee Q) \wedge (\neg P \vee \neg Q)$

$ = P \leftrightarrow \neg Q$.

Hence B is the answer.
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using x--> y <=> ¬ x V y
then ( x--> y ) $\Lambda$ ( y --> x) <=> x <--> y
Answer:

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