Natural deduction proof of $(\alpha\to\beta)\to(\beta\to\gamma)\to(\alpha\to\gamma)$ My teacher has assigned us this exercise as part of our homework:

Give a natural deduction proof of $(\alpha\to\beta)\to(\beta\to\gamma)\to(\alpha\to\gamma)$

Here is an example of natural deduction that he gave us:

$\vdash \alpha\to\beta\to\alpha$.
This is a tautology, but it's easier to prove than to verify:
$\alpha,\beta\vdash\alpha$, hypothesis
$\alpha\vdash\beta\to\alpha$, deduction theorem
$\vdash\alpha\to\beta\to\alpha$, deduction theorem

Here is my attempt:

Hypothesis: $(\alpha\to\beta),(\beta\to\gamma) \vdash (\alpha\to\gamma)$
Deduction thm: $(\alpha\to\beta) \vdash(\beta\to\gamma)\to(\alpha\to\gamma)$
Deduction thm: $\vdash(\alpha\to\beta)\to(\beta\to\gamma)\to(\alpha\to\gamma)$

Is this enough? I feel like I need to take apart the functions inside the parentheses and check those as well, but I don't know how.
 A: $$\begin{align}
 (1) & \alpha \rightarrow \beta && [\text{HYP}] \\
 (2) & \beta  \rightarrow \gamma && [\text{HYP}] \\
 (3) & \alpha && [\text{HYP}] \\
 (4) & \beta && [\text{MP}(1,3)] \\
 (5) & \gamma && [\text{MP}(2,4)] \\
 (6) & \alpha \rightarrow \gamma && [\rightarrow\text{-intro}(3,5)] \\
 (7) & (\beta \rightarrow \gamma) \rightarrow (\alpha \rightarrow \gamma) && [\rightarrow\text{-intro}(2,6)] \\
 (8) & (\alpha \rightarrow \beta) \rightarrow ((\beta \rightarrow \gamma) \rightarrow (\alpha \rightarrow \gamma)) && [\rightarrow\text{-intro}(1,7)] \\
\end{align}$$
A: Unless you already have that hypothesis, you need to get to that hypothesis first.
Also, (α→β)→(β→γ)→(α→γ) is ambiguous.  You almost surely want to prove ((α→β)→((β→γ)→(α→γ))).   Additionally, you don't actually use the deduction (meta) theorem, you use the rule of inference that it implies as valid.  I'll call this rule "conditional introduction".
I don't know your format exactly or what you have exactly, but a proof might go something like this:
Hypothesis (rule of modus ponens/detachment): (α→β), α $\vdash$ β
Weakening: (α→β), α, (β→γ) $\vdash$ β
Commutation on the left side: (α→β), (β→γ), α $\vdash$ β
Identity:  (α→β), (β→γ), α $\vdash$ β, (β→γ)
Detachment: (α→β), (β→γ), α $\vdash$ γ
Conditional introduction: (α→β), (β→γ) $\vdash$ (α→γ) 
A: I will replace $\alpha$ with $A$, $\beta$ with $B$, and $\gamma$ with $C$ to use the truth table generator and Fitch-style natural deduction proof checker.
To resolve the question of ambiguity, the following truth table shows that one cannot prove $((A\to B)\to(B\to C))\to(A\to C)$:

The counterexample is in the fourth line of the table where $A=T$, $B=F$, and $C=F$.
With the ambiguity resolved, here is a natural deduction proof:


Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
A: If, as Doug suggested, you are really required to prove
$A\implies B \implies (B\implies C \implies (A\implies C))$, 
you could start with 3 successive premises, the first two being:


*

*Suppose $A\implies B$

*Suppose  $B\implies C$

*Suppose ???
(I don't like Greek letters!)
