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].
$ \lnot (A \Rightarrow B) \vDash A \land \lnot B $

Remember that we can write $(A \Rightarrow B)$ as $\lnot A \lor B$.

So we have $\lnot(\lnot A \lor B)$, and distributing the $\lnot$, we get $A \land \lnot B$, as wanted.