The statement pattern [(p vee q) wedge sim p] | vedclass

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a contradiction

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(A) To determine the nature of the statement pattern,we construct a truth table:| $p$ | $q$ | $\sim p$ | $\sim q$ | $p \vee q$ | $(p \vee q) \wedge \sim p$ | $[(p \vee q) \wedge \sim p] \wedge \sim q$ ||---|---|---|---|---|---|---|| $T$ | $T$ | $F$ | $F$ | $T$ | $F$ | $F$ || $T$ | $F$ | $F$ | $T$ | $T$ | $F$ | $F$ || $F$ | $T$ | $T$ | $F$ | $T$ | $T$ | $F$ || $F$ | $F$ | $T$ | $T$ | $F$ | $F$ | $F$ |Since all entries in the final column are $F$ (False),the statement pattern is a contradiction.

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

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