How to prove that $(\neg q\implies p)\implies(p\implies \neg q)\equiv (\neg p\lor \neg q)$

Question

How to prove that $(\neg q\implies p)\implies(p\implies \neg q)\equiv \neg p\lor \neg q$ ?

I have come up that $(\neg q\implies p)\implies (p\implies \neg q)\equiv \neg(q\lor p)\lor\neg (p\land q)$ but I don't know how to go down to " $\equiv\neg p\lor\neg q$ ".

Answer 1

\begin{equation} (\neg q \implies p) \implies (p \implies \neg q) \\ \neg (\neg q \implies p) \lor (p \implies \neg q) \\ \neg(q \lor p) \lor (\neg p \lor \neg q) \\ (\neg q \land \neg p) \lor (\neg p \lor \neg q) \\ (\neg q \lor \neg p \lor \neg q) \land (\neg p \lor \neg p \lor \neg q) \\ (\neg p \lor \neg q) \land (\neg p \lor \neg q) \\ \neg p \lor \neg q \end{equation}

Answer 2

Note that for the LHS of your equation $v(\neg(q\lor p)) =0 \iff v(q)=v(p)=1$ (here $v$ stands for truth value).

Similarly for the RHS - $v(\neg(p\land q)) =0 \iff v(p)=v(q)=1$.

$ \therefore v(p)=v(q)=1 \iff v(\neg(q\lor p)\lor\neg(p\land q)) =0$. The contrary case is thus satisfied when

$v(p)=0\lor v(q)=0 \iff v(\neg p)=1\lor v(\neg q)=1$.

$\therefore \neg(q\lor p)\lor\neg(p\land q)\equiv(\neg p\lor \neg q)$.