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8. Let P and Q be two propositions, ¬ (P ↔ Q) is equivalent to :

(1) P ↔ ¬ Q

(2) ¬ P ↔ Q

(3) ¬ P ↔ ¬ Q

(4) Q → P

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3..?
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can u explain how option 3 @  MiNiPanda

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p<->q means (p->q)•(q->p)

= (!p+q)(!q+p) = p xnor q.

From the options given you can check that option 1 and 2 will give (p xor q) and last one will give (!q+p). Only option 3 is giving pxnorq.

$¬(P\leftrightarrow Q)$

$¬(P\rightarrow Q \wedge Q\rightarrow P)$

$¬ ((\bar{P}\vee Q) \wedge (\bar{Q}\vee P))$

$¬(\bar{P}\bar{Q} \vee PQ)$

$¬(P \odot Q)$

$(P\oplus Q)$

now check which options looks like XOR

P $\bar{P}$ Q $\bar{Q}$ $\bar{P} \leftrightarrow Q$ $P \leftrightarrow \bar{Q}$
T F T F F F
T F F T T T
F T T F T T
F T F T F F

$\bar{P} \leftrightarrow Q$ and $P \leftrightarrow \bar{Q}$ both are correct

by Boss (36.7k points)
edited

+1 vote