TITLE: Hilbert System Logical Axiom 1 follows from Axioms 2 and 3
QUESTION [5 upvotes]: I'm reading Wikipedia and it lists the first four logical axioms that allow (together with modus ponens) for the manipulation of logical connectives.

$\phi \to \phi $
$\phi \to \left(\psi \to \phi \right)$
$\left(\phi \to \left(\psi \rightarrow \xi \right)\right)\to \left(\left(\phi \to \psi \right)\to \left(\phi \to \xi \right)\right)$
$\left(\lnot \phi \to \lnot \psi \right)\to \left(\psi \to \phi \right)$

Then it states "The axiom 1 is redundant, as it follows from 3, 2 and modus ponens."
I see that if I substitute (2) into (3), I get
$$\left(\phi \to \left(\psi \rightarrow \phi \right)\right)\to \left(\left(\phi \to \psi \right)\to \left(\phi \to \phi \right)\right)$$
Since (2) is true, modus ponens tells me
$$\left(\phi \to \psi \right)\to \left(\phi \to \phi \right)$$
So (1) would be true if I knew that $\left(\phi \to \psi \right)$ is true. But how do I know that?
If there's a better way to do this proof than the way I approached it, I'll also accept that answer.

REPLY [6 votes]: Hint: Remember you get to pick what $\psi$ is.  Is there any formula $\psi$ such that you know $(\phi\to\psi)$ is true?
More details are hidden below.

 By axiom 2, if you choose $\psi$ to have the form $(\psi'\to\phi)$ for some formula $\psi'$, then you know $(\phi\to \psi)$ is true.

REPLY [4 votes]: By (3) we have
$$
(\phi\to((\phi\to\phi)\to\phi))\to((\phi\to(\phi\to\phi))\to(\phi\to\phi))
$$
By (2),
$$
\phi\to ((\phi\to\phi)\to\phi)
$$
So
$$
((\phi\to(\phi\to\phi))\to(\phi\to\phi))
$$
by Modus Ponens.
By (2),
$$
\phi\to (\phi\to\phi)
$$
so
$$
\phi\to\phi
$$
by Modus Ponens again.