$A \subset S$ then $A\cap S =A$ How would I prove the following
$A \subset S$ then $A\cap S =A$
Not sure I been thinking of an example say $A={1,2,3}$ and $S={1,2,3,4,5,6}$
So I tried the following
$(x:x\in A \wedge x\in S)$
so I let $x:x\in A$ I am not sure if this finishes the proof. Because I can can also $x\in S$
I think it would also be true if $(x:x\in S)$

I would say it's pretty easy to show this through a Venn diagram, if it helps you.
Also, if you have a set S for which $A \subset S$, then any $a\in A$ is also in S, but not every $s\in S$ is in A. So, you have:
$$A\cap S = \{s\in S | s \in A\} = \{s \in A\} = A $$
I think that's enough if you just want to understand why that is, but for more detail look here: https://proofwiki.org/wiki/Intersection_with_Subset_is_Subset
