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Commutative Law of Binary Operation

Without using...

Question

Without using truth table show that
$\sim (p \lor q) \lor (\sim p \land q) \equiv\sim p$

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Solution

We have,
L.H.S. $=\sim (p \lor q) \lor (\sim p \land q)$
$= (\sim p \land \sim q) \lor (\sim p \land q)$ ....(By De Morgan's Law)
$=\sim p \land (\sim q \lor q)$ ....(By Distributive Law)
$=\sim p \land T$ ....(By Complement Law)
$=\sim p =$ R.H.S.

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