not Z ⊆ not Y implies Y ⊆ Z by contradiction
not Z ⊆ not Y implies X ∩ Y ⊆ X ∩ Z (using part 1)
So far for 1 I have if not Z is subset of not Y then all elements of not Z are in not Y which means all elements not Z must not be in Y which means all elements of Y must be in Z. Is this correct? And how would I go about part 2?
EDIT: not Z would mean the complement of Z within an universal set U
$\endgroup$ 21 Answer
$\begingroup$That's not bad but is not "by contradiction".
If Y is not a subset of Z then there exist some y that is in Y but is not in Z. So y is in "not Z" but is not in "not Y". That contradicts the fact that "not Z" is a subset of "not Y".
For (2), if "not Z" is a subset of "not Y" then, by 1, Y is a subset of Z. If $x\in X\cap Y$ then x is in X and x is in Y. Since Y is a subset of Z, x is in Z. Since x is in X and in Z it is in $X\cap Z$.
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