The statement [(p wedge q) rightarrow p] righ | vedclass

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

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contradiction

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

The statement $[(p \wedge q)$ $\rightarrow p]$ $\rightarrow (q \wedge \sim q)$ is

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