Consider an effectively enumerable theory $T$ . Now, $\sigma\in T$ iff $T\vDash\sigma$ . And there is a procedure that can construct a sequence $\Sigma=\{\sigma_{0},\sigma_{1},\sigma_{2},\ldots\}$ of all sentences in $T$ . $\Sigma$ is a set of axioms of $T$ , and it is effectively enumerable, but we need a decidable set of axioms. As suggested, consider the set of axioms $\Sigma'=\{\sigma_{0},(\sigma_{0}\wedge\sigma_{1}),((\sigma_{0}\wedge\sigma_{1})\wedge\sigma_{2}),\ldots\}$ .$\mathfrak{A}$ is a model of $\Sigma$ iff it is a model of $\Sigma'$ (which is quite straightforward to see). But now, $\Sigma'$ is decidable. Indeed, having an input sentence $\sigma$ (if it is not a sentence, we reject it), we just check the length of $\sigma$ , and see, if there is a sentence in $\Sigma'$ having the same length as $\sigma$ by generating a necessary number of sentences (every next sentence in $\Sigma'$ has the length larger than the previous one), and if there is one, we check whether it is $\sigma$ or not.