Prove that $A \cup (M\setminus{A})=M$ Ok. I need to prove that:
$$
A \cup (M\setminus{A})=M
$$
I did that like this:
$$
A \cup (M\setminus{A})=M\\
x\in A \lor (x\in M \land x\notin A)=x \in M\\
(x\in A \lor x\in M )\land (x\in A \lor x\notin A)=x \in M\\
x\in A \lor x\notin A \;\text{is tautology, always true}\\
x\in A \lor x\in M =x \in M\\
$$
And this is the part that is confussing for me, because I cant say that $x\in A \land x\in M =x \in M\\$, because what if $x\in A \lor x\notin M =x \in M\\$. Then I still get true as a expression, but $x\notin M$. Can I say the that the expression is not valid?
 A: You can. For example if $A=\{1,2\}$, $M=\{2,3\}$ then
$$A\cup (M \setminus A) = \{1,2\}\cup\{3\}=\{1,2,3\} \ne M$$
A: You can indeed say that the given "pseudo-identity" is not a valid identity. All you need to prove this is to provide a counterexample, as you suggested: find two simple example sets such that $x\in A$ but $x\notin M$, and you'll have proven that the expression is not valid for all sets, $A, M$.
Note, however, if we add the restriction that $A\subseteq M$, then $\;A \cup (M\setminus A) = M\;\;\text{is valid}.$
A: Assuming you have $A \subseteq M$, let $x\in A\cup (M\backslash A)$. Then:
$$
\begin{align}
\notag x\in A\cup (M\backslash A) &\Rightarrow x\in A \vee x\in M\backslash A
\end{align}
$$
Case 1: $x\in A$: If $x\in A$ then $x$ is obviously in $M$, because $A \subseteq M$. 
Case 2: $x\in (M\backslash A)$. Then $x$ is obviously in $M$. Thus $A\cup (M\backslash A) \subseteq M$. Can you show $M \subseteq A\cup (M\backslash A)$? 
Note: If $M\subseteq A$, then $M\backslash A = U\backslash A$, where $U$ is the universal set. - Correct me if I'm wrong, I'm an beginning analysis student also.
