If $\Gamma \cup \{ \neg \varphi \}$ is inconsistent, then $\Gamma \vdash \varphi$

Question

If $\Gamma \cup \{ \neg \varphi \}$ is inconsistent, then $\Gamma \vdash \varphi$

Here, a set of formulas is inconsistent means they syntactically imply some formula as well as its negation. Syntactic implication here consists of (1) given formulas, (2) all first-order logical axioms, (3) modus ponens.

I have tried to explicitly construct a syntactic proof for the conclusion but failed to come up with one.

Answer 1

Suppose $\Gamma\cup\{\neg\alpha\}$ is inconsistent. Then for some $\beta$ we have $\Gamma\vdash\beta$ and $\Gamma\vdash\neg\beta$. It remains only to show that $\{\beta,\neg\beta\}\vdash\alpha$.

Answer 2

You need the axiom (or tautology) :

$\vdash (¬φ→¬ψ)→((¬φ→ψ)→φ)$.

Having already proved :

if $Γ∪{¬φ}$ is inconsistent, then $Γ⊢¬φ→ψ$ and $Γ⊢¬φ→¬ψ$,

it is enough to aplly modus ponens twice to conclude with :

$Γ⊢φ$.