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I have to make or deny proof for : $\exists X \forall Y \exists Z .(X\subseteq Y \cap Z) $

Is the proof by saying that $\emptyset \subseteq All$ and so $X = \emptyset$
enough?

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    $\begingroup$ Could you state in english what the statement you are trying to prove is? It sounds like you are trying to prove: There exists a set $X$ so that for every set $Y$ there exists a set $Z$ so that $X\subseteq Y\cap Z$ which is... as you argue ... trivial. But a bizarre statement (who cares about the $Z$?) $\endgroup$ – fleablood Nov 29 '19 at 17:29
  • $\begingroup$ yeah.. It sound very trivial.. In fact, it's not mine, I found it in one exercise book online and it just seems so easy, that I'm confused a little bit .. There is nothing more than this the exercise gave me.. $\endgroup$ – Patrik Bašo Nov 29 '19 at 17:30
  • $\begingroup$ Well, Sure $X=\emptyset $ will be a subset of $Y\cap Z$ for any $Y$ and any $Z$. And it's the only such set the statement is true. If $X$ is non empty then there is an $x \in X$ an there can exists sets $Y$ so that $x \not \in Y$ and so $X \not \subset Y$ and nor can $X\subset W\subset Y$ for any subset, $W$, of $Y$. So there can not exist any set $Z$ so that $X\subset Y\cap Z$.... I guess it's a valid excercise in concept and definitions .... but basic. You seem to be beyond the level of such definition quiz cards. $\endgroup$ – fleablood Nov 29 '19 at 17:37
  • $\begingroup$ Yeah. exactly. I must find some more difficult ones $\endgroup$ – Patrik Bašo Nov 29 '19 at 17:39
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Yes it is, there is no trick question.

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