Getting weird results

Question

Consider two nonempty sets $ S $ and $ T $, $S\subseteq T$. We can write: $$\forall x (x\in S \Rightarrow x\in T) $$ Knowing, that $ p\Rightarrow q $ is equivalent to $\neg (p \wedge \neg q) $, we can rewrite the previous expression as: $$\forall x (\neg (x\in S \wedge x\notin T)) $$ Turning the universal quantifier into the existential one: $$\neg (\exists x (x\in S \wedge x \notin T)) $$ Splitting the quantifier into two: $$\neg (\exists x (x\in S)\wedge \exists x (x\notin T)) $$ The first conjuct in parentheses is given to be true, so we rewrite the expression again: $$\nexists x (x\notin T) $$ Or, more familiarly: $$\forall x (x\in T) $$ Which is a complete nonsense. Where is the mistake in reasoning here?

Answer 1

You can't split the quantifier as you did. That's where the problems begin.

$$\neg (\exists x (x\in S \wedge x \notin T))\not\equiv \neg (\exists x (x\in S)\wedge \exists x (x\notin T)) $$

The existentially quantified statement $\exists x (x \in S \land x \notin T)$ asserts that there exists some $x$ for which two things are true about that one specific $x$, $P(x): x\in S$ and $Q(x): x \notin T$.

When you split the clause into two existentially quantified statements, you are asserting merely that there exists something, say $x$, for which $P(x)$ is true, and there exists something (perhaps something else, we'll call it $y$) for which $Q(y)$ is true. Those aren't equivalent statements.

Answer 2

To expand a little on the comment, while we have $$\exists x(x\in S\wedge x\notin T)\implies\exists x(x\in S)\wedge\exists x(x\notin T)$$

the converse is false, so these are not equivalent statements. Adding the $\neg$ reverses the direction of the implication, so we get:

$$\neg(\exists x(x\in S\wedge x\notin T))\implies\neg((\exists x(x\in S)\wedge\exists x(x\notin T))$$

Therefore your "proof" only works in reverse, which is fine; if $T$ is the entire universe, then every other set in that universe is a subset of it.