If the truth value of the statement pattern $[\mathrm{p} \wedge \sim \mathrm{r}] \rightarrow \sim \mathrm{r} \wedge \mathrm{q}$ is False, then which of the following has truth value False?

Given the statement $S = [p \wedge \sim r] \rightarrow (\sim r \wedge q)$ with truth value False, determine which of the provided options also evaluates to False.

An implication $A \rightarrow B$ is false only when $A$ is true and $B$ is false. Therefore:

$p \wedge \sim r$ must be true, which requires both $p$ and $\sim r$ to be true, so $p \equiv T$ and $r \equiv F$.

Since $\sim r \wedge q$ must be false and $\sim r \equiv T$, it follows that $T \wedge q$ is false only if $q \equiv F$.

Truth values: $p \equiv T$, $q \equiv F$, $r \equiv F$.

Evaluating the options:

Option A: $(p \vee r) \rightarrow \sim r$ becomes $(T \vee F) \rightarrow T$, which simplifies to $T \rightarrow T$ and is true.

Option B: $(r \vee q) \rightarrow \sim p$ becomes $(F \vee F) \rightarrow F$, or $F \rightarrow F$, which is true.

Option C: $\sim(p \vee q) \rightarrow \sim r$ becomes $\sim(T \vee F) \rightarrow T$, or $F \rightarrow T$, which is true.

Option D: $\sim(r \vee q) \rightarrow \sim p$ becomes $\sim(F \vee F) \rightarrow F$, or $T \rightarrow F$, which is false.

Only option D has truth value False.

Final answer: $\boxed{D}$