Given that A, B and C are sets, and that (A ∪ B) \ C ⊆ A \ B
(A cup B) minus C (is contained in) A minus B
Prove that (A \ C) ∩ B = ∅
(A minus C) cap B = Empty set
I tried to prove it this way:
It is given that the containing set (A \ B) doesn't contain any $x \in B$ (by the definition of set difference). Therefore, no $x \in $ B && x $\notin$ C exists. Therefore, in no way there is x $\in$ B && x $\notin$ C && x $\in$ A exists (equivelent to (A \ C) ∩ B = ∅ )
I'm afraid my proof is incorrect, because $x \notin A$ seems unneccesary.
I'm a new student in the university in Israel, learning parallel to my highschool studies. English isn't my mother tongue. I am sorry for any mistakes and my bad formatting.
(Added by A.K. translation of the Hebrew in the image)
By the assumption every $x$ which belongs to $A$ or $B$, and does not belong to $C$ is necessarily an element of $A$ and not an element of $B$ (by the definitions of inclusion, union and difference). In the right hand side ($A\setminus B$) no element belongs to $B$, therefore it is impossible that in the left hand side there is an element which belongs to $B$. Therefore there is no $x\in B$ and $x\notin C$. In particular there is no such $x$ for which $$x\in A\land x\in B\land x\notin C$$ Q.E.D