= Solution
The following is a <logical derivation> using only A1, A2 and the <modus ponens rule>. The first two lines are assumptions:
$$
\begin{array}{rll}
1.&\alpha\to\beta&\text{assumption},\\
2.&\beta\to\gamma&\text{assumption},\\
3.&(\beta\to\gamma)\to(\alpha\to(\beta\to\gamma))&\mathrm{A1},\\
4.&\alpha\to(\beta\to\gamma)&\mathrm{MP}\ (2,3),\\
5.&(\alpha\to(\beta\to\gamma))\to((\alpha\to\beta)\to(\alpha\to\gamma))&\mathrm{A2},\\
6.&(\alpha\to\beta)\to(\alpha\to\gamma)&\mathrm{MP}\ (4,5),\\
7.&\alpha\to\gamma&\mathrm{MP}\ (1,6).
\end{array}
$$
Thus $\boxed{\{\alpha\to\beta,\beta\to\gamma\}\vdash\alpha\to\gamma}$.
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