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Let ${\lnot}$, ${\in}$ and ${\implies}$ be undefined notions.

Then, in the language of set theory, $p{\land}q{\iff}{\lnot}(p{\implies}{\lnot}q)$, where $p$ and $q$ are some WFFs.

Let $p$ be $a\,{\in}\,{\{a,b,c,d,e}\}$.

Let $q$ be ${\lnot}{\exists}{\emptyset}:{\lnot}{\exists}a\,{\in}\,{\emptyset}$.

$p{\land}q$ is also understood as "p is true and q is true".

Since ${\emptyset}$ described in $q$ does exist, $q$ is false.

It isn't true that $p$ implies not $q$, so by the first definition $p{\land}q$.

However, since $p$ is true and $q$ is false, then by the second definition ${\lnot}(p{\land}q)$.

How do we resolve this?

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    $\begingroup$ But p does imply not q. $\endgroup$ – fleablood Apr 4 '16 at 20:17
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Your $q$ misuses notation rather badly. What you mean is that $q$ is $\neg\exists x\,\forall y\,\big(\neg(y\in x)\big)$, i.e., there is no set that has no elements. As you say, this is false. Thus, $\neg q$ is true, and $p\to\neg q$ is true no matter what statement $p$ is. Of course this means that $\neg(p\to\neg q)$ is false no matter what $p$ is, and so is $p\land q$.

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But it is true that $p \implies \lnot q $ ($p $ and $\lnot q$ are true). So $\lnot (p \implies \lnot q) $ and therefore $p\land q $ is false.

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