For what three sets $A,B,C$ is this equality true: $(A \cap B)\times(B\cup C) = (A\times B) \cup (A\times C) $? For what three sets $A,B,C$ is this equality true: $(A \cap B)\times(B\cup C) = (A\times B) \cup (A\times C) $? Justify your answer.
I've struggling with this for a while, but I'm quite lost. Could you help me out how to approach this type of problems?
 A: $(A \cap B)\times(B\cup C) = (A\times B) \cup (A\times C)  
= A×(B \cup C)$.  
If A is empty, then B and C can be anything.
If B is empty, then A×C is empty.
If B is not empty, then $(A \cap B) = A$. 
A: First, you should try to find a condition where it works : Here you will notice quite easily that your equality holds if $A \subset B$.
Then try to look what can happen if your condition isn't verified. If your condition isn't the most general one, looking at other situations will help you to notice it and give you hints at what you should be looking for. Else if nothing else works, you can try to prove that your condition is the right one, but this time you know what you are trying to prove which makes it easier. In our case :
Suppose $A$ isn't a subset of $B$. So $\exists a \in A\setminus B$. Then if either $B$ or $C$ aren't empty, let $v\in B\cup C$ : $(a,v)\in A\times (B\cup C) = (A\times B) \cup (A\times C)$ but $(a,v)\notin (A \cap B)\times (B\cup C)$, so the equality doesn't hold.
If both $B$ and $C$ are empty, then the equality holds since $A\times \emptyset = \emptyset$.
So in our case everything worked out well by trying to see what happens when our condition didn't held. We have proved that this equality was true if and only if $A \subset B$ or $B= C = \emptyset$.
