The statement (p rightarrow q) rightarrow [(s | vedclass

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

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a tautology

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

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

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