Prove $\{\phi\} \vdash \psi$ if and only if $\vdash (\phi \rightarrow \psi)$

Question

I'm self-teaching myself some logic, and I was hoping that someone can check my proof of the statement in the title, and give comment on whether this is formatted acceptably.

(This problem comes from the textbook I'm learning from, Mathematical Logic by Chiswell and Hodges, Ex. 2.4.5).

The rules I use, as declared in the textbook, are:

($\rightarrow I$): $\Gamma \cup \{\phi\} \vdash \psi \Rightarrow \Gamma \vdash (\phi \rightarrow \psi)$

($\rightarrow$ E): $(\Gamma \vdash \phi) \wedge (\Delta \vdash (\phi \rightarrow \psi)) \Rightarrow (\Gamma \cup \Delta \vdash \phi)$

(Axiom Rule): $\phi \vdash \phi$

And the proof is:

1. Prove $\{\phi\} \vdash \psi \Leftrightarrow~ \vdash (\phi \rightarrow \psi)$
2. $\{\phi\} \vdash \psi \Rightarrow~ \vdash (\phi \rightarrow \psi)$ (from $\rightarrow$ I).
3. $\vdash (\phi \rightarrow \psi) \Rightarrow \{\phi\} \vdash \psi$ --(From Axiom Rule and $\rightarrow$ E)
4. It is proven because 2. and 3. and definition of $a \Leftrightarrow b \equiv a \Rightarrow b \wedge b \Rightarrow a$

I'm not sure if I should be elaborating further on each step, or if this is how proofs are usually formatted. Any feedback would be welcomed.