Disproving with a counter example when changing which side is the subset

Question

I originally wanted to know how to disprove by counter example and found out that I was trying to disprove something that was true.

MPW and Brian in the comment section of MPW's answer helped me discover that while it might be true that one side is a subset of the other, the reverse may not be true.

Answer 1

Suppose $x\in (A\cap B\cap C)\setminus D$. Then $x\in A$, $x\in B$, $x\in C$, and $x\notin D$.

• Since $x\in B$, certainly $x\in B\cup C$.

• Since $x\in B\cup C$ and $x\notin D$, we know $x\in(B\cup C)\setminus D$.

• Since $x\in A$ and $x\in(B\cup C)\setminus D$, we know $x\in A\cap ((B\cup C)\setminus D)$.

We have just shown that if $$x\in (A\cap B\cap C)\setminus D$$ then $$x\in A\cap ((B\cup C)\setminus D)$$ In other words, we have shown that $(A\cap B\cap C)\setminus D \subseteq A\cap ((B\cup C)\setminus D)$.

So the statement is in fact true. You won't find any counterexample since non exists.

Answer 2

Drawing a Venn diagram could be an idea. It's a bit difficult with four sets, though. Here is one way (it's easy, but not necessarily pretty, and might technically not be a Venn diagram, but it gets the job done): Draw a Venn diagram for $A,B,C$, then inside each of the eight regions, draw a circle. $D$ is the union of all these circles.

Now you can just check which of the 16 regions are contained in which of your two sets. This is basically the most general setting you can have. If a Venn diagram says that your inclusion is true, then will always be true, as a Venn diagram is an abstract representation of any possible way sets can overlap. If it shows that your inclusion is false, then you have a concrete counterexample right in front of you.