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

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Question

(A) The given expression is $(p \wedge \sim q) \wedge (\sim p \vee q)$.We construct the truth table to evaluate the expression:      $p$    $q$    $\s

Vedclass · Accepted Answer

$A$ contradiction

Vedclass · Answer

(A) The given expression is $(p \wedge \sim q) \wedge (\sim p \vee q)$.We construct the truth table to evaluate the expression:      $p$    $q$    $\sim p$    $\sim q$    $p \wedge \sim q$    $\sim p \vee q$    $(p \wedge \sim q) \wedge (\sim p \vee q)$        $T$    $T$    $F$    $F$    $F$    $T$    $F$        $T$    $F$    $F$    $T$    $T$    $F$    $F$        $F$    $T$    $T$    $F$    $F$    $T$    $F$        $F$    $F$    $T$    $T$    $F$    $T$    $F$  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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