$Q$ is a relation as described above, $\Sigma$ is consistent, and $\phi$ is a formula with one variable. I think the relation in above holds because if $c$ does not belong in $Q$, then by the relation $\Sigma$ cannot prove $\phi[c]$. Not being able to prove a sentence is not the same as proving the negation of the sentence, right?
Also, what if I have another formula $\psi(w)$ and I'm given $Q(c) \iff \Sigma \vdash \lnot \psi[c]$? Does $\lnot Q(c) \iff \Sigma \not\vdash \lnot\psi[c]$? I'm thinking it is so because I treat $\lnot \psi$ as $\phi$ and argue like before.