The negation of statement pattern $(p \wedge \sim q) \rightarrow(p \vee \sim q)$ is

Finding the negation of $S = (p \wedge \sim q) \rightarrow (p \vee \sim q)$ reveals that it is a contradiction.

The negation of an implication $A \rightarrow B$ is $A \wedge \sim B$, so $\sim S \equiv (p \wedge \sim q) \wedge \sim (p \vee \sim q)$.
Applying De Morgan's law, $\sim (p \vee \sim q) \equiv \sim p \wedge q$, resulting in $\sim S \equiv (p \wedge \sim q) \wedge (\sim p \wedge q)$.

Rearranging terms yields $(p \wedge \sim p) \wedge (\sim q \wedge q)$, which simplifies to $F \wedge F$ since a proposition conjoined with its negation is false.
Therefore, $\sim S \equiv F$, confirming it is a contradiction.

Alternatively, verifying that $S$ is a tautology also shows its negation must be a contradiction.

Thus, the negation of the statement pattern is indeed a contradiction.