Prove that $f$ is recursive given that $T\vdash\phi(a,b)$ iff $f(a)=b$ Prove that $f$ is recursive given that $T\vdash\phi(\bar{a},\bar{b})$ iff $f(a)=b$ and $T$ is an axiomatizable theory of arithmetic, where $\bar{x}$ is the numeral in the language of arithmetic $L_A$ for the number $x$.
My reasoning thus far is as follows:

*

*Since $T$ is axiomatizable we have: $\Gamma\vdash\phi(\bar{a},\bar{b})$ iff $f(a)=b$ for some $\Gamma\subseteq T$

*For any system $\vdash\phi(\bar{a},\bar{b})$ iff $\vdash\exists x\exists y(x=\bar{a}\land y=\bar{b}\land\phi(x,y))$

*So, $\Gamma\vdash\exists x\exists y(x=\bar{a}\land y=\bar{b}\land\phi(x,y))$ iff $f(a)=b$

*The set $F^+$ of pairs of numbers $(n,m)\in\mathbb{N}$ such that $\Gamma\vdash\exists x\exists y(x=\bar{a}\land y=\bar{b}\land\phi(x,y))$ is semirecursive.

*For any consistent system $\vdash\psi$ iff $\nvdash\lnot\psi $

*So, $\Gamma\vdash\exists x\exists y(x=\bar{a}\land y=\bar{b}\land\phi(x,y))$ iff $\Gamma\nvdash\lnot\exists x\exists y(x=\bar{a}\land y=\bar{b}\land\phi(x,y))$

*The set $F^-$ of pairs of numbers $(n,m)\in\mathbb{N}$ such that $\Gamma\nvdash\lnot\exists x\exists y(x=\bar{a}\land y=\bar{b}\land\phi(x,y))$ is semirecursive.

*Using Kleene's Theorem 4 and 7 imply that $F^+$ is recursive.

*And so $f$ is recursive. Since $F^+ = f$
I have serious misgivings about step 7. But I have been banging my head against the wall for the last 48 hours on this one to no avail. Any and all help is greatly appreciated. For reference this is exercise 15.10 in Boolos, Burgess, and Jeffrey's Computability and Logic.

It is easy to see that the graph of $f$ is semidecidable. To see that also its complement is semidecidable (and therefore, that $f$ is decidable), observe that $f(a)\neq b$ iff $f(a)=c$ for some $c\neq b$. Hence $f(a)\neq b$ iff there exists $c\neq b$ such that $T\vdash\varphi(\bar{a},\bar{c})$, and this latter property is clearly semidecidable.
