Negation of $$p \wedge (q \wedge \sim(p \wedge q))$$ is

We need the negation of $$p \wedge (q \wedge \sim(p \wedge q))$$.

First, we simplify the inner part: $$q \wedge \sim(p \wedge q) = q \wedge (\sim p \vee \sim q)$$. Distributing gives $$= (q \wedge \sim p) \vee (q \wedge \sim q) = (q \wedge \sim p) \vee F = q \wedge \sim p$$. Substituting back into the full expression yields $$p \wedge (q \wedge \sim p) = (p \wedge \sim p) \wedge q = F \wedge q = F$$.

Because the expression is always false, its negation is always true.

To find a matching tautology, consider Option 1: $$(\sim(p \wedge q)) \vee p$$. This simplifies as follows: $$(\sim(p \wedge q)) \vee p = (\sim p \vee \sim q) \vee p = (\sim p \vee p) \vee \sim q = T \vee \sim q = T$$, showing it is a tautology.

The correct answer is Option 1: $$(\sim(p \wedge q)) \vee p$$.