Let $S=\{x\in\Bbb Z:\exists y\in\Bbb Z:x=24y\}$ and $T=\{x\in\Bbb Z:\exists y,z\in\Bbb Z:x=4y\land x=6z\}$. Prove that $S\ne T$.
Here's what I've done so far. I know that $S$ is a subset of $T$, but $T$ is not a subset of $S$. So $S$ cannot equal $T$. I am having trouble proving $T$ is not a subset of $S$.
suppose $x$ is in the set of $T$. Therefore $x$ is divisible by $4$ and $6$. How do I finish off this proof?
Edit: Consider $24 = 4y$ and $24 = 6z$. $y = 6$ while $z = 4$. $6$ can be rewritten as $2y' = 6$, while $2z' = 4$. This is the definition of even numbers.
I then get $y' = 3, z' = 2$. Now I substitute those values into the original equation. $x = 4(3)$ and $x = 6(2)$. both resulting in $12$. Have I shown in a valid way that the set $T$ contains factors of $12$, while set $S$ does not?
If you have to show two sets are not equal, it suffices to find an element in one set and not the other. In this case, $12\not\in S$ but $12\in T$.
It turns out that $S\subsetneq T$, but this is not needed for the question at hand.