proving standard logic equivalences

Question

Trying to prove if this is logically equivalent

$p \leftrightarrow (q \rightarrow r)$ and $(p \leftrightarrow q) \rightarrow r$

I got that they wouldn't be equal just wanted to confirm if I was right. My final answer was $((p\lor r)\land (\neg q\lor r))\land (\neg r\lor (\neg p\lor q))$ after simplifying the right hand side.

Answer 1

Examine $\def\getsto{{\;\leftrightarrow\;}}\def\false{\underset{\tiny\text{false}}\bot}\def\true{\underset{\tiny\text{true}}\top} p \getsto (q \to r)$ and $(p \getsto q) \to r$

$\begin{align}p \getsto (q \to r) ~& \iff (\neg p\vee (q\to r)\wedge(p\vee\neg(q\to r)) \\ &\iff (\neg p\vee\neg q\vee r)\wedge(p\vee\neg(\neg q\vee r))) \\ &\iff (\neg p\vee\neg q\vee r)\wedge (p\vee(q\wedge\neg r)) \\[3ex](p \getsto q) \to r ~&\iff \neg(p\getsto q)\vee r \\ & \iff (p\oplus q)\vee r \\ & \iff ((\neg p\vee \neg q)\wedge(p\vee q))\vee r \\ & \iff (\neg p\vee \neg q\vee r)\wedge(p\vee q\vee r) \end{align}$

Test when $r$ true but $p,q$ both false.

$ \false\getsto(\false\to\true)$ is false but $(\false\getsto\false)\to\true$ is true.