Does this equivalency proof work for $\lnot (A \rightarrow B) \equiv A \wedge \lnot B$?

Question

I had a question about one of my proofs... it seemed right to me, but I didn't know if it would be generally acceptable:

The question is to prove, using equivalencies, the following: $\lnot(A \rightarrow B) \equiv A \wedge \lnot B$

$\equiv \hspace{1cm}\lnot (\lnot A \vee B)\hspace{1.4cm}imp\\ \equiv \hspace{1cm}\lnot (\lnot A) \wedge \lnot B \hspace{1cm}DeMorgan\\ \equiv \hspace{1cm} A \wedge \lnot B \hspace{2cm}imp$

I was wondering if this would work, or if I had to do some other step?

Any help would be appreciated. Thank you.

Answer 1

The steps are okay, but the third step is justified by double negation equivalence.

$$\begin{align}&\neg(A\to B)\\\equiv~& \neg(\neg A\vee B)&&\text{Implication}\\\equiv~&\neg\neg A\wedge \neg B&&\text{deMorgan}\\\equiv~&A\wedge\neg B&&\text{Double Negation}\end{align}$$