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...?
 A: 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].

A: Your proof is so far so good!
So, to continue from:
$$\neg A \text{ Assumption}$$
$$\text{new box}$$
$$A \text{ Assumption}$$
$$\bot \text{ (from A and not A)}$$
$$B \text{ (from contradiction you can infer anything)}$$
$$\text{end box}$$
$$A \rightarrow B$$
$$\bot$$
$$\text{end box}$$
$$\neg \neg A$$
$$A$$
$$A \land \neg B$$
A: If you assume $\neg A$ you should be able to prove that $A\Rightarrow B$. This can be done by deduction theorem, because if you assume $A$ you can prove $B$ by contradiction (and by deduction theorem you get $\neg A\vdash A\Rightarrow B$). Now with the assumption $\neg(A\Rightarrow B)$ you have a contradiction and can conclude that the assumption $\neg A$ was false or that $A$ is proven.
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.
A: What you have to do is show that $\lnot (\lnot(A\to B)\to(A\land\lnot B))$ is a tautology. But the tableau of its negation is closed:
.
(Each path contains a contradiction.) Hence the result.
