(p wedge sim q) wedge (sim p vee q) is :- | vedclass

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(A) To determine the nature of the statement,we construct a truth table:$p$$q$$\sim p$$\sim q$$(p \wedge \sim q)$$(\sim p \vee q)$$(p \wedge \sim q) \

Vedclass · Accepted Answer

$A$ contradiction

Vedclass · Answer

(A) To determine the nature of the statement,we construct a truth table:$p$$q$$\sim p$$\sim q$$(p \wedge \sim q)$$(\sim p \vee q)$$(p \wedge \sim q) \wedge (\sim p \vee q)$$T$$T$$F$$F$$F$$T$$F$$T$$F$$F$$T$$T$$F$$F$$F$$T$$T$$F$$F$$T$$F$$F$$F$$T$$T$$F$$T$$F$Since the final column contains only $F$ (False) for all possible truth values of $p$ and $q$,the statement is a contradiction.

$(p \wedge \sim q) \wedge (\sim p \vee q)$ is :-

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