If $A\subset B$ and $C\subset D$, then $A\cap C \subset B\cap D$ and $A'\cap C'\subset B'\cap D'$ Let $A$, $B$, $C$, and $D$ be subsets of $U$ (the universe) with $A$ as a subset of $B$ ($A \subset B$) and $C$ as a subset of $D$ ($C \subset D$). Is it always true that:


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*The intersection of $A$ and $C$ is a subset of the intersection of $B$ and $D$?

*The intersection of what's not in $A$ or $C$ is a subset of the intersection of what's not in $B$ or $C$? 
 A: We have $A\subseteq B$ and $C\subseteq D$.


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*Yes, $A\cap C\subseteq B\cap D$. The standard approach to proving a result like this is what I call element-chasing: let $x$ be an arbitrary element of $A\cap C$, and prove that $x$ must necessarily belong to $B\cap D$. So suppose that $x\in A\cap C$. Then $x\in A\subseteq B$, so $x\in B$, and $x\in C\subseteq D$, so $x\in D$. In other words, $x\in B$ and $x\in D$, so by definition $x\in B\cap D$. We used no information about $x$ beyond the assumption that $x\in A\cap C$, so it follows that every element of $A\cap C$ belongs to $B\cap D$, which by definition means that $A\cap C\subseteq B\cap D$.

*Following the comment, I’m interpreting this as $A'\cap B\,'\subseteq B\,'\cap C'$. Suppose that we try to prove this in the same way. Let $x\in A'\cap B\,'$; then $x\in A'$ and $x\in B\,'$, and we want to show that $x\in B\,'$ and $x\in C'$. We have $x\in B\,'$, so that’s no problem, but it turns out that we can’t guarantee that $x\in C'$. What if $C=D=U$? Then $C'=\varnothing$, so $$B\,'\cap C'=B\,'\cap\varnothing=\varnothing\;.$$ It’s still possible that $A'\cap B\,'\subseteq\varnothing$, but only if $A'\cap B\,'=\varnothing$, and we can surely construct an example in which that’s not the case. Here’s a very simple one: take $C=D=U=\{0\}$ and $A=B=\varnothing$. Then $A\subseteq B$, $C\subseteq D$, and $$A'\cap B\,'=\{0\}\cap\{0\}=\{0\}\nsubseteq\varnothing=\{0\}\cap\varnothing=B\,'\cap C'\;.$$
In connection with the second problem you should try to prove that if $A\subseteq B$, then $B\,'\subseteq A'$, so $A'\cap B\,'=B\,'$, and we could have simplified the inclusion to $B\,'\subseteq B\,'\cap C'$; it’s probably even easier to suspect that this isn’t always true.
