Equivalent characterisation of consistent

Question

We call a set of formulas $\Sigma$ of a language $L$ consistent if there is no $\varphi$ in $L$ such that $\Sigma \vdash \varphi$ and $\Sigma \vdash \lnot \varphi$.

Apparently, an equivalent formulation is the following:

A set $\Sigma$ of formulas of $L$ is consistent iff $\Sigma \not\vdash \varphi$ for some sentence $\varphi$ of $L$.

The $\implies$ direction is clear: if we can prove all sentences then we can prove both $\varphi$ and $\lnot \varphi$ so that $\Sigma$ is inconsistent.

But I don't immediately see how to prove $\Longleftarrow$. Can someone explain this to me? Thanks!

Answer 1

Assume that $\Sigma$ is consistent then it cannot prove $\varphi\land\lnot\varphi$. Therefore there exists a sentence which it does not prove.

Assume that $\Sigma$ is inconsistent then it proves everything (using the principle of explosion). Therefore there is no sentence it does not prove.

Answer 2

The characterizations are not equivalent in general. Take the most trivial case: suppose $L$ is a negation-free language. Then vacuously, the set $\Sigma$ of all $L$-formulae is consistent in the first sense, but not in the second sense.

However, in a language with negation and in the presence of ex contradictione quodlibet (the rule that from $\varphi$ and $\neg\varphi$ you can derive any $\psi$), the two characterizations become co-extensional.