If both $\Phi, \neg\phi$ and $\Phi, \psi$ are inconsistent, then $\Phi \vdash \neg(\phi \to \psi)$.

Question

Could someone check whether my solution is okay?

If both $\Phi, \neg\phi$ and $\Phi, \psi$ are inconsistent, then $\Phi \vdash \neg(\phi \to \psi)$.

Answer 1

You need to specify that you are assuming the arbitrary valuation function, $\sigma$, shall hold $\Phi$ to be true.

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Take as given that $\Phi,\neg\phi$ and $\Psi,\psi$ are inconstistent.   Assume that a valuation function, $\sigma$, satisfies $\Phi$ .   That is $\sigma(\Phi)=\mathrm T$.   Therefore $\sigma(\neg\phi)\neq\mathrm T$ and $\sigma(\psi)\neq\mathrm T$ because of the inconsistancies.   In classical logic that is $\sigma(\phi)=\mathrm T$ and $\sigma(\psi)=\mathrm F$.   From that we infer $\sigma(\phi\to\psi)=\mathrm F$, since $\phi\to\psi$ cannot be satisfied when $\phi$ is true but $\psi$ is false.   Therefore there is no evaluation that satisfies both $\Phi$ and $\phi\to\psi$, that is $\Phi,\phi\to\psi$ is inconsistent.   Consequently $\Phi\vDash \neg(\phi\to\psi)$ .