# Prove that $\left( A-B\right) \cup B = A \cup B$

To prove:$\left( A-B\right) \cup B = A \cup B$
I want it to be done in two ways:1. The algebraic way and 2 Using the method where we say,for example, $x \in A \text{ or }x\notin B$(I dont know what that method is called.)
I have done the algebraic method, somebody help me do it using the second method where we assume that let x be a element that belongs to LHS and then we prove that LHS is a subset of the RHS. Then we do it all again but with y such that it belongs to RHS and finally we prove that RHS is a subset of LHS.

• What are your thoughts on the matter? What have you trid? What sort of proof are you looking for - an algebraic proof like the one you wrote below, or a prose proof emphasizing quantifiers? – Carl Mummert Nov 14 '13 at 13:55
• @CarlMummert The OP already answered the question – Amr Nov 14 '13 at 13:56
• @CarlMummert I posted this question earlier here:math.stackexchange.com/questions/566763/… but was asked by lord_farin to make a separate question for it. – Shaurya Gupta Nov 14 '13 at 14:00
• @Amr: yes, but I don't think that is relevant to the issue of the question being written in a way that is not yet suitable for this site. Here is a link to "How to ask a good question": meta.math.stackexchange.com/questions/9959/… – Carl Mummert Nov 14 '13 at 14:03
• @CarlMummert I think it's relevant. Some users post questions and answer it, but I believe their questions are interesting/non-trivial. I consider this question is trivial. It is uninteresting to see a lot of trivial questions on the site. I know trivial questions are being asked, but at least the posters don't know the answer, but I don't think its a good idea to tolerate trivial questions being asked by users who know their answer – Amr Nov 14 '13 at 14:07

Let $x\in (A-B)\cup B$, then $x\in A-B$ or $x\in B$. If $x\in A-B$, then $x\in A$ and $x\notin B$ and so $x\in A\cup B$. If $x\in B$, then $x\in A\cup B$. Hence $(A-B)\cup B\subset A\cup B$. Let $x\in A\cup B$, then $x\in A$ or $x\in B$. If $x\in A$, then $x\in A\cup B$. If $x\in B$, then $x\in A\cup B$. Hence $A\cup B \subset (A-B)\cup B$. Thus $(A-B)\cup B=A\cup B$.
$\left( A-B\right) \cup B = (A \cap B') \cup B = (B \cup A) \cap (B \cup B') = (A \cup B) \cap X = A \cup B$
Note that $A,B \in X$