I was reading these notes and the definition of $\Sigma-$representability of a relation. It's as followings. $R$ is $\Sigma$ representable if $\forall a \in \mathbf N^m$ (where $a_i = S^{a_i} 0$ and $S$ is the successor function):
- $R(a) \implies \Sigma \vdash \varphi(a)$
- $\neg R(a) \implies \Sigma \vdash \neg \varphi(a)$
however for some reason the definition does not apply both ways automatically (which is what I would have expected). It does when $\Sigma$ is consistent (for some reason).
- $R(a) \iff \Sigma \vdash \varphi(a)$
- $\neg R(a) \iff \Sigma \vdash \neg \varphi(a)$
Why does that only happen when $\Sigma$ is consistent (and thus can be completed easily by Lindenbaum).
If $\Sigma$ is inconsistent, it proves everything, so $\Sigma\vdash \varphi(\mathbf a)$ and $\Sigma \vdash \lnot \varphi(\mathbf a)$ always hold, whereas at most one of $R(a)$ or $\lnot R(a)$ hold, so it is impossible for both $\leftarrow$ directions to hold for any $a,$ let alone all of them.