If $\Phi \vdash \phi \to \psi$, show that $\Phi, \phi, \neg\psi$ is inconsistent. 
If $\Phi \vdash \phi \to \psi$, show that $\Phi, \phi, \neg\psi$ is inconsistent.

I am stuck on my proof.

Assume that $\Phi \vdash \phi \to \psi$. Then $\Phi, \neg(\phi \to \psi)$ is inconsistent or $\Phi, \phi \wedge \neg \psi$ is inconsistent. Then $\sigma(\phi \wedge \neg \psi) = F$.

I am not sure where to go from here since the falsity can be made in three ways and I just wanted one way, where $\sigma(\phi)=\sigma(\neg\psi)=F$.
 A: Use the deduction theorem.
$\Phi\vdash \phi\to\psi$, so $\Phi,\phi\vdash\psi$.  Hence $\Phi,\phi,\lnot\psi$ syntactically entails both $\psi$ (coming from $\Phi,\phi$) and $\lnot\psi$ (from $\lnot\psi\vdash\lnot\psi$), so is inconsistent.
Or with valuations:
Let $v$ be a valuation that assigns $\top$ to all of $\Phi,\phi,\lnot\psi$.  From $\Phi\vdash\phi\to\psi$, we have $v(\phi\to\psi)=\top$.  Also, by supposition $v(\phi)=\top$, so we must have $v(\psi)=\top$, contradicting $v(\lnot\psi)=\top$.
A: 
Assume that $\Phi \vdash \phi \to \psi$. Then $\Phi, \neg(\phi \to \psi)$ is inconsistent or $\Phi, \phi \wedge \neg \psi$ is inconsistent. 

If you are allowed to use those substitution equivalences, you're done.
When $\{x, y\land z\}$ is inconsistent, so too $\{x,y,z\}$ is inconsistent, because $\nu(y\land z)=\mathrm T$ iff $\nu(y)\land\mathrm T$ and $\nu(z)\land\mathrm T$. 
So $\{x, y, z\}$ cannot be satisfied by any evaluation exactly when $\{x, y\land z\}$ cannot be satisfied.

...

But more simply, use deduction theorem: When $\Phi\vdash \phi\to \psi$, then we may infer that $\Phi, \phi\vdash \psi$ and so...
