let $\alpha$ be a sentence (closed formula) in number theory language, such that $PA\nvdash\neg\alpha$. For any sentence $\beta$ , let us denote $g(\beta)$ as $\beta$'s Godel number. We define the next sets:
$S= \{ g(\beta) : PA \vdash \alpha \rightarrow\beta \}$
$T= \{ g(\beta) : PA \vdash \beta \rightarrow\alpha \}$
$T\cup S$
When $PA$ denotes Peano's axioms. I wish to demonstrate that $S,T,S\cup T$ are not recursive.
For S: One can use deduction theorem: $S= \{ g(\beta) : PA\cup \alpha \vdash \beta \}$, we know that $PA\nvdash \neg\alpha$ so $PA\cup\alpha$ is consistent, therefore $PA\cup \alpha$ is not complete by Godel's theorem, so there is a formula $\gamma$ such that $PA\cup\alpha \nvdash \gamma$ and $PA\cup\alpha \nvdash \neg\gamma$ .
Another approach, is using the fact that $cons(PA)$ is not recursive, maybe we may assume that $cons(PA\cup\alpha)$ is recursive, and by that defining an algorithm (we accept Church thesis) to construct an algorithm which determines for every $\gamma$ whether $\gamma \in cons(PA)$ which contradicts $PA$ undecidability?