Proving $ \lnot (A \Rightarrow B) \vDash A \land \lnot B $ I need help trying to prove
$$
   \lnot (A \Rightarrow B) \vDash A \land \lnot B
$$
in natural deduction.
I've come this far:
$$ \begin{array}{|l}\hline
     \lnot (A \Rightarrow B) \text{ premise} \\
     ~~\begin{array}{|l}\hline
       B \text{ assumption} \\
       ~~\begin{array}{|l}\hline
         A \text{ assumption} \\
         B \text{ copy} \\\hline
       \end{array}\\
       A \Rightarrow B \\
       \bot \\\hline
     \end{array}\\
     \lnot B \\
     ~~\begin{array}{|l}\hline
       \lnot A \text{ assumption}\\\vdots\\\hline
     \end{array}\\\hline
   \end{array}
$$
And then...?

1) $\lnot (A \to B)$ --- premise
2) $\lnot (A \land \lnot B)$ --- assumed [a]
3) $A$ --- assumed [b]
4) $\lnot B$ --- assumed [c]
5) $A \land \lnot B$ --- from 3) and 4) by $\land$-intro
6) $\bot$ --- from 2) and 5)
7) $B$ --- from 4) and 6) by Double Negation-elim, discharging [c]
8) $A \to B$ --- from 3) and 7) by $\to$-intro, discharging [b]
9) $\bot$ --- from 1) and 8)

10) $A \land \lnot B$ --- from 2) and 9) by Double Negation-elim, discharging [a].

