Prove if true and find a counterexample if false (disproofs/algebraic proofs) For all sets $A, B, C$
$$A- (B-C) = (A-B) -C$$
Now my books says it is false and begins showing a counter example, but how do you know it is false by just looking at it? Would you go about proving it first and then if a proof can't be found, make up a counter example?
The proof I think would start out...
$x$ in $A$ and not ($x$ in $B$ and $x$ not in $C$)=
 A: There are several possibilities:


*

*You have enough intuition to immediately see that it looks unlikely to be true, so try to build a counterexample.

*You try drawing some simple examples (in this case probably in the form of Venn diagrams) to see if you accidentally find a counterexample. If it turned out that the statement was true, this method might also give you some clues as to why.

*The statement looks similar to other statements that you know are false. There isn't anything extremely obvious in your example, although if you replaced the sets by numbers and the set subtractions by usual subtraction, the resulting expression would be false (i.e. subtraction of integers is non-associative). This is obviously not a proof that your statement is false, but maybe makes it look less plausible.

*You mistakenly decide the statement is true, and try to prove it. Your proof inevitably breaks down, but possibly in such a way as to tell you what a counterexample looks like.
There are probably more things you can do, but these are all good strategies for dealing with "prove or find a counterexample" type questions.
A: In this particular case I would focus on $C$. It’s immediately clear that $(A\setminus B)\setminus C$, is disjoint from $C$, i.e., contains no points of $C$. On the left, $B\setminus C$ contains no points of $C$, and it’s what I’m subtracting from $A$, so I’m not removing any points of $C$ from $A$. If $A$ contains some point $c$ of $C$, then $A\setminus(B\setminus C)$ is still going to contain that point, but $(A\setminus B)\setminus C$ isn’t.
Then I try to build a specific counterexample based on this idea. I might let $A=C=\{c\}$ and $B=\varnothing$, just to keep things simple. Does it work?
$$A\setminus(B\setminus C)=\{c\}\setminus(\varnothing\setminus\{c\})=\{c\}\setminus\varnothing=\{c\}\;,$$
and 
$$(A\setminus B)\setminus C=(\{c\}\setminus\varnothing)\setminus\{c\}=\{c\}\setminus\{c\}=\varnothing\;,$$
so it works exactly as expected.
I could also have done a quick check of ‘extreme’ special cases. What happens if $A=\varnothing$? Nothing useful: both sides are $\varnothing$. What if $B=\varnothing$? Then
$$A\setminus(B\setminus C)=A\setminus(\varnothing\setminus C)=A\setminus\varnothing=A\;,$$
and
$$(A\setminus B)\setminus C=(A\setminus\varnothing)\setminus C=A\setminus C\;.$$ 
Certainly those don’t have to be equal; for instance, I could take $A=C\ne\varnothing$, and they would be unequal.
