How to prove that finding math proofs is $\text{NP}$-hard?
Define $\mathrm{Bool}$ to be the set of all subsets of $\{\varnothing\}$ (or some arbitrary set with 1 element). Then represent a propositional variable $x$ by some subset $x\in\mathrm{Bool}$, i.e. $x\subseteq\{\varnothing\}$.

Now there is an extremely straightforward translation $\rho$ of propositional formulae into such subsets:


$\rho(x)$ if $x$ is a variable is just some set (also called $x$) $\in\mathrm{Bool}$.
$\rho(\neg\phi)$  into the complement of $\rho(\phi)$ in $\mathrm{Bool}$.
$\rho(\psi\vee\phi)=\rho(\psi)\cup\rho(\phi)$
$\rho(\phi\wedge\psi)=\rho(\psi)\cap\rho(\phi)$


And you don't need any other connectives to get a polynomial sized translation of a propositional formula.

Then a formula $\phi$ is satisfiable iff
$$ ZF\vdash\exists x_1\ldots x_n\in\mathrm{Bool},\ \rho(\phi) \neq \varnothing$$

I'll let you work out the details. One important thing, though, is to show that a proof will be proportional in length to $n$, where $n$ is the number of variables.