I am studying discrete mathematics and i found an exercise that asks to prove or disprove that , if A,B,C are sets then : (A intersection with C) = (B intersection with C) implies A = B. The solution i found online was the counter example C={null set}. The exercise is from Rosen's Book 8th edition , page 145 , exercise 32 , question b.
My proof: Suppose (A intersection with C) = (B intersection with C) is true (1). A = B iff (A subset of B) and (B subset of A). Let's show first that (A subset of B). We need to show that if x belongs to A then it belongs to B. Suppose x belongs to A and (1) is true , then we conclude: (x belongs to C) implies (x belongs to B and x belongs to C) Therefore , we cannot conclude that belongs to B because we don't know if x belongs to C or not. So , A is not necessarily subset of B , and it is not the case that A=B.
My thought process is that sometimes is hard if not impossible to find a counter example. And if i use logic i can reach to a statement in which i can't prove the conclusion of the implication. Therefore the conclusion is incorrect without additional premises.
Is this a correct way to disprove this ? Or you must find a counterexample ?
No, your proof is incorrect. Inability to reach certain conclusion tells us nothing whether the conclusion is true or not, because you not being able to do something doesn't mean that it is impossible.
This is a common mistake and to avoid it you first need to identify whether your problem asks you to prove something for all sets or that there exist some sets for which it is true.
Let me consider cases of what could happen.
In this case you need to produce certain deduction that starts from "let $x$ be a set" and through a sequence of valid inferences (due to axioms or known theorems) ends up at statement $P(x)$.
In this case all you need to do is find one such $x$. We'd call such an $x$ an example to $P$.
To prove that the above statement is false, we prove that the negation of the statement is true and the negation is "There exists $x$ such that not $P(x)$." Notice that this is actually the same form as 2., so we want to find $x$ such that $P(x)$ is not true. We'd call such an $x$ a counterexample to $P$.
Like previous example, we want to prove the negative, which in this case is "For all $x$: not $P(x)$", which is of the same form as 1.
Your particular problem is of the form for all, so we want to do either case 1. or case 3. In practice when we problem solve, we won't know which case we need, whether we need to prove that statement is true or prove that the negative of the statement is true. A good strategy is often to start writing a proof as if we were in case 1. If we can write proof, great. If we can't write proof, we need to identify why exactly we can't write proof. You actually did that, you wrote: "Therefore , we cannot conclude that belongs to B because we don't know if x belongs to C or not."
At that point, you realized you are actually not in case 1., but in case 3. instead. When that happens, you always need to produce counterexample, as written above. Now, the sentence you wrote that I quoted is actually very useful to find counterexample, you realized the following: if you could guarantee that $x$ is in $C$, the proof would be done, and if you could guarantee that $x$ is not in $C$, you'd get a counterexample. The easiest way to do it is to pick $C$ to be the empty set, since it gives counterexample for any set $A$. Other $C$ would work as well, but you'd have to think about $A$ as well (see the comment by Jaap Scherphuis).