Show that if, for some $ψ$, $Γ \vdash ψ$ and $Γ \vdash ¬ψ$, then $Γ \vdash φ$, for any formula $φ$.

Question

Show that if, for some $ψ$, $Γ \vdash ψ$ and $Γ \vdash ¬ψ$, then $Γ \vdash φ$, for any formula $φ$. [Hint: Use a result from Exercise 3.7, as attached below.]

(Could anyone check my proof please? Also would it matter if I use g instead of f? Because while g is a tautology and f isn't, they both look very similar, if not, almost identical.)

Suppose $Γ \vdash ψ$ and $Γ \vdash ¬ψ$, and we have them on $k$ and $k+1$ lines respectively. Then we utilise Ex3.7f to modus ponens and produce $φ$. Since one can substitute $φ$ with any formula, we can - as asked by the question - derive any formula $φ$.

$$ \begin{matrix} \vdots & \vdots & \vdots \\ (k) & ψ & ... \\ (k+1) & ¬ψ & ... \\ (k+2) & ¬ψ\vdash(ψ\toφ) & 3.7f \\ (k+3) & (ψ\toφ) & (k+1),(k+2) MP \\ (k+4) & φ & k,(k+3) MP \\ \end{matrix} $$