The following is a problem from an old qualifying exam in logic:
Let $\mathcal{L}$ be a finite language consisting only of function symbols. Show that then the set of universal $\mathcal{L}$-sentences $\sigma$ such that $\models \sigma$ is decidable.
An earlier part of the same problem asked for a proof of the following result: If $\Sigma$ is a decidable set of $\mathcal{L}$-sentences such that, for each $\sigma \in \Sigma$, we have $\models \sigma$ if and only if $\mathcal{A} \models \sigma$ for each finite $\mathcal{L}$-structure $\mathcal{A}$, then $\left\{\sigma \in \Sigma : \, \models \sigma \right\}$ is decidable. My plan of attack is to use this result by defining $\Sigma$ to be the set of universal $\mathcal{L}$-sentences (in which case $\Sigma$ is decidable) and showing that, if $\mathcal{A} \models \sigma$ for each finite $\mathcal{L}$-structure $\mathcal{A}$, then $\models \sigma$. However, I have not been able to prove that $\mathcal{A} \models \sigma$ for each finite $\mathcal{A}$ implies $\models \sigma$. Assuming there are arbitrarily large finite $\mathcal{L}$-structures, I know from the compactness theorem that there is at least one infinite $\mathcal{L}$-structure satisfying $\sigma$; why must all infinite $\mathcal{L}$-structures satisfy $\sigma$? And, what is the importance of the assumption that $\mathcal{L}$ consists only of function symbols?