I have little experience with math proofs and I would like to prove $A \cup (A\cap B)=A$, by showing that the left hand side is a subset of right hand side and vice versa.
Since $ A \subseteq A \space and \space A\cap B \subseteq A$
$\therefore \space lhs \subseteq rhs$
Given that $A \cup (A\cap B) \equiv A \cap (A \cup B)$
$A \subseteq A \space and \space A \subseteq (A\cup B) $
$\therefore \space rhs \subseteq lhs$
Does the above proposed solution suffice as a proof?
$\endgroup$ 21 Answer
$\begingroup$The question you ask is quite subjective. Suffice for whom? In what context? If you are a freshman then your proofs are judged in different standards then the proofs of a second- or third-year student.
It also depends on what have you proved before (e.g. $X\subseteq Y$ and $Z\subseteq Y$ implies that $X\cup Z\subseteq Y$).
Note that the for $\text{RHS}\subseteq\text{LHS}$ you only need to note that $A\subseteq A$ and therefore $A\subseteq A\cup X$ for every $X$, in particular $X=A\cap B$.
You may also wish to write an element chasing proof, e.g. to show $A\subseteq A\cup(A\cap B)$:
$\endgroup$Let $a\in A$ be an arbitrary element, we will show that $a\in A\cup(A\cap B)$. Since $a\in A$ we have that $a\in A$ or $a\in A\cap B$. Therefore $a\in A\cup(A\cap B)$, and therefore $A\subseteq A\cup(A\cap B)$.
