2

Let $\Gamma$ be a maximal consistent set. Prove: $\varphi \lor \psi \in \Gamma \iff \varphi \in \Gamma $ or $ \psi \in \Gamma$.

Now define $V_{\Gamma}: Q \to \{ 0, 1 \}$ as follows:

$V_{\Gamma}(p):= \cases{{1 \mbox{ if } p \in \Gamma} \\ 0 \mbox{ if } p \notin \Gamma}$.

Show that for every formula $\varphi$ we have: $\varphi \in \Gamma \iff V_{\Gamma} \models \varphi$.

I'm at a loss here. Could anyone please help me out?

Thanks in advance.

Jeroen
  • 1,653
  • If you look up the definition of $V_\Gamma \models \phi$, I think you will be able to come up with an answer for the second part yourself. – Lord_Farin May 15 '13 at 17:57

1 Answers1

3

Here's an answer for the more difficult direction of the first part. Suppose, toward a contradiction, that $\Gamma$ contains $\phi\lor\psi$ but contains neither $\phi$ nor $\psi$. Since it doesn't contain $\phi$ but is maximal consistent, $\Gamma\cup\{\phi\}$ must be inconsistent. So $\neg\phi$ is a logical consequence of $\Gamma$. Similarly, so is $\neg\psi$. But then so is $\neg(\phi\lor\psi)$ because that's a logical consequence of $\{\neg\phi,\neg\psi\}$. But since $\phi\lor\psi$ is in $\Gamma$, this contradicts the assumption that $\Gamma$ is consistent.

Andreas Blass
  • 71,833