Show that $(\phi \rightarrow \psi), (\phi \rightarrow \neg \psi) \vdash \neg \phi$

In your previous post you have proved Double Negation elimination :

$\vdash (¬¬p → p)$

with axioms 1 and 3; so we assume that we can use it.

We assume also the Proof by contradiction theorem :

If $\Gamma, ¬φ \vdash ψ$ and $\Gamma, ¬φ \vdash ¬ψ$, then $\Gamma \vdash φ$.

Proof

1) $(φ → ψ)$ --- assumed

2) $(φ → ¬ψ)$ --- assumed

3) $\vdash \lnot \lnot φ → φ$ --- DN

4) $(φ → ψ), \lnot \lnot φ \vdash ψ$ - from 3) by modus ponens

5) $(φ → ¬ψ), \lnot \lnot φ \vdash ¬ψ$ - from 3) by modus ponens

Now we apply Thinning and then Proof by contradiction to 4) and 5) with $\Gamma = \{ (φ → ψ), (φ → ¬ψ) \}$ to derive :

6) $ (φ → ψ), (φ → ¬ψ) \vdash \lnot φ$.