1)Can we write a paradox in propositional logic?

2)I'm trying to solve the following problem : Let $\epsilon$ be a set of evaluations and $\gamma$ = {$\psi \in Form$ | $ e \models \psi $ for each evaluation in $\epsilon$ } ( $Form = $ the set of all formulae). We suppose there are at least 2 elements in $\epsilon$. Show that there exists a formula $\phi$ so that $\phi \notin \gamma$ and $\neg\phi \notin \gamma $ .

I was thinking that a paradox might be the answer? If not can someone show me a solution? I would greatly appreciate your help!

  • $\begingroup$ Seems like an a punctuation emergency. $\endgroup$ – copper.hat Feb 8 '17 at 23:48

Since you have at least two evaluations in $\epsilon$, let's consider two of those evaluations, $e_1$ and $e_2$. Since these are different, it must be true that for some proposition $p$: $e_1(p) \not = e_2(p)$ (that is, one evaluation will set that statement $p$ to True, while the other sets it to False). So, for this very statement $p$ it will not be true that for all evaluations $e \in \epsilon$: $e \vDash p$, so $p \not \in \gamma$. However, if $e_1(p) \not = e_2(p)$, then of course $e_1(\neg p) \not = e_2(\neg p)$, and so likewise it will not be true that for all evaluations $e \in \epsilon$: $e \vDash \neg p$, so $\neg p \not \in \gamma$.

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