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


  
    $Γ⊢φ$.
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 $\{\beta,\neg\beta\}\vdash\alpha$.