Use mathematical logic to prove this proposition

[solakis1]

Given the following axioms:
1) $P\implies(Q\implies P)$
2) $((P\implies(Q\implies R))\implies((P\implies Q)\implies(P\implies R))$ Where $P,Q,R$ are any formulas
3)$(\neg P\implies\neg Q)\implies (Q\implies P)$ then prove:

$\{A\implies B,B\implies C\}|- A\implies C$
Without using the deduction theorem and as a rule of inference M.ponens

 

[andrewkirk]

Your OP outlines the axioms of a Hilbert System. Go to the wiki page on Hilbert Systems and search "(HS2)" to see a proof of the following proposition from those axioms using Modus Ponens as rule of inference.
$$(p \to q) \to ((q \to r) \to (p \to r))$$
Relabel $p,q,r$ as $A,B,C$ to get
$$(A \to B) \to ((B \to C) \to (A \to C))$$
Then we have:
\begin{align}
&\vdash(A \to B) \to ((B \to C) \to (A \to C))\\
(A \to B), (B \to C)&\vdash(A \to B) \to ((B \to C) \to (A \to C))\\
(A \to B), (B \to C)&\vdash A\to B\quad\quad\textrm{[1st axiom]}\\
(A \to B), (B \to C)&\vdash(B \to C) \to (A \to C)
\quad\quad\textrm{[Modus Ponens on 3, 2]}\\
(A \to B), (B \to C)&\vdash B\to C \quad\quad\textrm{[2nd axiom]}\\
(A \to B), (B \to C)&\vdash A \to C
\quad\quad\textrm{[Modus Ponens on 5, 4]}
\end{align}