NOTE: The $F$ represents false.
I have this question in my hw, the thing is that I think that equivalence is not true.
Because $r \land ( \neg p \rightarrow F) \equiv r \land ( p \lor F) \equiv r \land p$
And by doing the truth table of $r \land p$ and $ r \rightarrow p$ they are not equal then I concluded they are not equivalent, but I would like to have your opinion, probably I am not seeing something important.
Let $r$ and $p$ be false. Then $(\lnot p)\to F$ is false${}^\dagger$, so $r\land ((\lnot p)\to F)$ is false. But $r\to p$ is true.
$\dagger$ I have just realised that this step is unnecessary, since, if $r$ is false, then $r\land ((\lnot p)\to F)$ is false. Sorry!
Perhaps you have the parentheses in the wrong place? My guess is that you're supposed to show that $(r \wedge \neg p) \to F$ is equivalent to $r \to p$.
