Prove or disprove $p\wedge(q\rightarrow r) \Longleftrightarrow (p\rightarrow q)\rightarrow r$ using logical equivalences

Question

Here is what I have so far:

\begin{align*} p \wedge (q \to r) & \iff p\wedge(\neg q \vee r) \\ & \iff (p \wedge \neg q) \vee (p \wedge r) \\ &\iff \neg(p \to q) \vee (p \wedge r) \\ &\iff (p \to q) \to (p \wedge r) \\ &\iff (p \to q) \to r \end{align*}

Could someone please verify whether this is correct, and how to improve it?

Answer 1

This is not a tautology, to disprove it, we only need to find a counter example.

Here we denote true as $1$, false as $0$.

Take $p=0,q=0,r=1$, then we have:

$$p ∧ (q → r) \equiv0\land(0\to1)\equiv0\land1\equiv0$$

$$(p → q) → r\equiv(0\to 0)\to1\equiv1\to1\equiv1$$

Since $1\neq0$, that '$p=0,q=0,r=1$' indeed is a counter example.

Hence this disproved the statement.