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I need to prove $A \subseteq B \Rightarrow A\cup (B\setminus A) = B$ This is what I did so far and I don't know where to go from there:

$x\in A\cup (B\setminus A) \Rightarrow x\in A \vee x\in(B\setminus A)$

And from definition of B\A I know : $x\in A \vee (x\in B \wedge x\notin A)$

From here I don't know where to go, I know from $A \subseteq B $ that

$x\in A \rightarrow x\in B$ which means $x\in B \vee (x\in B \wedge x\notin A)$

What to do from here ? is this expression the equivalent of $x\in B$ ? If so how ? If you can criticize my way before or instead of showing another direction to the question because I learn the best that way.

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    The key point is that the consequent $(A\cup (B\setminus A) = B)$ holds **if** $A\subseteq B$. If we do not assume $A\subseteq B$, there are counter examples $A, B$ for which $(A\cup (B\setminus A) \neq B)$ – 2017-01-01

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Well, you know that $x\in A \implies x\in B$, and $x\in B\setminus A \implies x\in B$, so $x\in A \cup (B\setminus A) \implies x\in B$. Thus, $A\cup(B\setminus A) \subset B$. Now, assume $x\in B$. Then $(x\in A) \lor (x\notin A \implies x\in (B\setminus A)$, so $B\subset A \cup (B\setminus A)$. Thus, the two sets are equal.

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As $A\subseteq B$ then, $A\cup B=B.$ Choosing $U=B$ as universal set: $$A\cup (B\setminus A)=A\cup(B\cap A^c)=(A\cup B)\cap (A\cup A^c)=B \cap U = B.$$

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    It's not really necessary to choose $U=B$, but otherwise this is definitely how I would have gone about the problem (avoid element-chasing whenever possible!). +1 – 2017-01-01
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    @Daniel W. Farlow: I agree. – 2017-01-01
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    @DanielW.Farlow Why is it not necessary ? How could you say $B \cap U = B$ otherwise? – 2017-01-01
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    @user3575645 What is $B\cap(B\cup X)$ where $X$ is any set whatever? In your problem, you deal with two sets, $A$ and $B$. What is the *smallest* the universal set could be? – 2017-01-01
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    @DanielW.Farlow $B\cap(B\cup X)$ is $B$, and I'm not sure what the smallest universal set that can be, but on a second thought isn't $B \cap U = B$ true by definition ? And because $ B \subseteq U $ always ( I think ) then $B \cap U = B$? – 2017-01-01