An exercise asks me to prove the following: $$A \cap (B-A) = \varnothing$$ This is what I did:
I need to prove that $A \cap (B-A) \subseteq \varnothing$ and $\varnothing \subseteq A \cap (B-A)$
The second one seems to be obvious by definition (not sure if it is fine to say that in a test).
But the first one goes like this: $$x \in A \cap (B-A)$$ $$x \in A \land x \in (B-A)$$ $$x \in A \land (x \in B \land x \notin A)$$ $$(x \in A \land x \notin A) \land x \in B$$ $$(x \in A \land x \notin A)$$ Since there is a contradiction... err... the proof... is good.
Okay, that's my problem. I think that the procedure is fine, but clearly I am unable to word it out (well, I don't even know if it is valid at all). Is it valid? How can I... "prove that my proof is correct"? I don't really see this contradiction as an obvious indicator of $A \cap (B-A) \subseteq \varnothing$