I want to show that $\vdash \phi \land \psi \leftrightarrow \psi \land \phi$ for any two formulas $\phi$ and $\psi$. Before I write down my thoughts, I would like to mention that I am by no means experienced with logic, so if I should mention what axioms I am assuming or anything else, please let me know (I remember the prof only saying that we are using a Hilbert deduction system and the deduction rule is modus ponens; I will also mention that this is an introductory logic course).
The exercise previously asked me to show the following four facts:
$$\{\phi \land \psi\} \vdash \phi$$
$$\{\phi \land \psi\} \vdash \psi$$
$$\{\phi, \psi\} \vdash \phi \land \psi$$
$$\{\phi, \psi\} \vdash \chi \text{ iff } \{ \phi \land \psi\}\vdash \chi \text{ for any formula }\chi$$
This is why I thought that some of them may come in handy. If I write the definition of $\leftrightarrow$ in terms of $\land$ and $\rightarrow$, I will end up with a horrendous expression, so this doesn't look like the way to go. I thought that I might use the completeness theorem and this is a valid way to solve the problem, but I think that I am supposed to solve it using syntactic results only.
EDIT: As requested, the axioms are:
$\bullet$ $\phi \rightarrow (\psi \rightarrow \phi)$
$\bullet$ $(\phi \rightarrow (\psi \rightarrow \chi)) \rightarrow ((\phi \rightarrow \psi)\rightarrow (\phi \rightarrow \chi))$
$\bullet$ $(\neg \psi \rightarrow \neg \phi) \rightarrow (\phi \rightarrow \psi)$,
where $\phi$, $\psi$ and $\chi$ are formulas.
Best Answer
As you have proved, $\{\phi,\psi\}\vdash\psi\land\phi$ (since $\{\phi,\psi\}=\{\psi,\phi\}$ and according to point 3 of your exercise). And according to point 4 this means that $\{\phi\land\psi\}\vdash\psi\land\phi$. Using the deduction theorem we gain $\vdash(\phi\land\psi)\rightarrow(\psi\land\phi).$ Similarly, $\vdash(\psi\land\phi)\rightarrow(\phi\land\psi)$.
Now it's quite simple to conclude $\vdash(\phi\land\psi)\leftrightarrow(\psi\land\phi).$