Negation of $$(p \to q) \to (q \to p)$$ is

To find the negation of $$(p \to q) \to (q \to p)$$ we use the fact that the negation of $$A \to B$$ is $$A \wedge \neg B$$, and here $$A = (p \to q)$$ and $$B = (q \to p)$$, so the negation becomes $$(p \to q) \wedge \neg(q \to p)$$.

Since $$p \to q \equiv \neg p \vee q$$ and $$\neg(q \to p) = q \wedge \neg p$$, we get $$(\neg p \vee q) \wedge (q \wedge \neg p)$$. Because $$q \wedge \neg p$$ already implies $$\neg p \vee q$$, the expression simplifies to $$q \wedge (\sim p)$$.

Therefore, the negation of $$(p \to q) \to (q \to p)$$ is $$q \wedge (\sim p)$$.