= Solution
First obtain $A$ by a <double-negation elimination>, then obtain $\neg B$ by the <negation introduction rule>. The following <natural deduction> proof uses only the printed rules, with $\bot E$ denoting the <logical falsity elimination rule>:
$$
\begin{array}{rll}
1&\neg(A\to B)&\text{premise}\\
2&[\neg A]^a&\text{assumption}\\
3&\quad[A]^b&\text{assumption}\\
4&\qquad\bot&\neg E\ 3,2\\
5&\qquad B&\bot E\ 4\\
6&\quad A\to B&\to I\ (3\text{--}5)\\
7&\quad\bot&\neg E\ 6,1\\
8&\neg\neg A&\neg I\ (2\text{--}7)\\
9&A&\neg\neg E\ 8\\
10&[B]^c&\text{assumption}\\
11&\quad[A]^d&\text{assumption}\\
12&\qquad B&\text{reiteration of 10}\\
13&\quad A\to B&\to I\ (11\text{--}12)\\
14&\quad\bot&\neg E\ 13,1\\
15&\neg B&\neg I\ (10\text{--}14)\\
16&A\land\neg B&\land I\ 9,15.
\end{array}
$$
Reiteration only reuses an available line within its scope. In particular, the unused <temporary assumption> on line 11 is legitimately discharged by the <implication introduction rule>.
$$
\boxed{\neg(A\to B)\vdash A\land\neg B.}
$$
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