I'm trying to prove $ M \subset N \Rightarrow M \cup(N \setminus M) = N $ I'm trying to prove:
$$
M \subset N \Rightarrow M \cup(N \setminus M) = N
$$
My trial:
$
\{x|x\in M \cup(N \setminus M)\} \Leftrightarrow \{x|x\notin \overline{M \cup(N \setminus M)}\}
$
using de Morgan
$ \Leftrightarrow \{x|x\notin \overline{M} \cap \overline{(N \setminus M)}\}$
$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \notin \overline{(N \setminus M)}\}$
$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \in (N \setminus M)\}$
$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \in N \wedge x\notin M\}$
aand we got a problem here. $x$ cannot be both $x\notin \overline{M}$ and  $x\notin M$.
My steps seems logical to me. Can someone please spot my mistake. Thanks a lot!
 A: Your error: $x\notin A\cap B$ is the negation of $x\in A\cap B$. $$x\in A\cap B \text{ iff }x\in A \land x\in B$$
So,$$x\notin A\cap B \text{ iff }x\notin A \lor x\notin B$$
Please notice how the $\land$ becomes $\lor$ on negation! In general, $\lnot(p\land q) = \lnot p\lor \lnot q$.

Alternative proof strategy: Whenever I want to show $A = B$, where $A,B$ are two sets, I aim to establish $A\subset B$ and $B\subset A$. Works on most occasions!

Suppose $M\subset N$. Then, for every $x\in M$, $x\in N$. Note that
$$N\setminus M = \{x\in N: x\notin M\}$$
Suppose $x\in M \cup N\setminus M$. Then, $x\in M$ or $x\in N\setminus M$. If $x\in M$, then $x\in N$ by $M\subset N$. If $x\in N\setminus M$, then $x\in N$ by definition of $N\setminus M$. So, $M\cup N\setminus M \subset N$.
Next, suppose $x\in N$. Then, either $x\in M$ or $x\notin M$. If $x\in M$, $x\in M\cup N\setminus M$ by definition of the union. If $x\notin M$, then clearly $x\in N\setminus M$, since we assumed $x\in N$, to begin with (refer to the definition of $N\setminus M$ again). As a result, $x\in M\cup N\setminus M$ by definition of the union, once again. So, we have $N\subset M\cup N\setminus M$.
$M\cup N\setminus M \subset N$ and $N\subset M\cup N\setminus M$ together, yield $M\cup N\setminus M = N$. So we have shown $$M\subset N\implies M\cup N\setminus M = N$$ Done!

Lastly, the following image says more than words ever can:

A: My advice is that you should prove each inclusion separately (in one you will have to use the hypothesis, but in the other you don’t need to).
Let $M$ and $N$ be sets. Suppose that $M \subseteq N$. We want to prove that $M \cup (N \setminus M) = N$. By definition of equality of sets, we must prove that
\begin{align*}
x \in M \cup (N \setminus M) \iff x \in X.
\end{align*}
Let’s prove each of the implications. Let $x \in M \cup (N \setminus M)$. Hence
\begin{align*}
x \in M \cup (N \setminus M) & \implies x \in M \vee x \in N \setminus M & (\text{definitions of union}) \\
& \implies x \in N \vee x \in N \setminus M & (\text{by hypothesis, }M \subseteq N) \\
& \implies x \in N \vee (x \in N \wedge x \notin M) & (\text{definition set difference}) \\
& \implies x \in N \vee x \in N & (\text{definition of conjunction})\\
& \implies x \in N & (\text{idempotent property of } \vee)
\end{align*}
which allows us to conclude that $M \cup (N \setminus M) \subseteq N$. Now, let $y \in N$. Then
\begin{align*}
y \in N & \implies y \in N \setminus M & (\text{definition of set difference}) \\
& \implies y \in M \vee y \in N \setminus M & (\text{addition}) \\
& \implies y \in M \cup (N \setminus M) & (\text{definition of set union})
\end{align*}
from which we conclude that $N \subseteq M \cup (N \setminus M)$.
Hence, we have that $N \subseteq M \cup (N \setminus M) \subseteq N$. Therefore, $M \cup (N \setminus M) = N$. $\square$
Any doubt on these, just let me know.
A: Your mistake :
$x \notin A \cap B \implies (x \notin A) \lor (x \notin B) $ not what you wrote. Think of it like this, $x$ not being in one of those keeps it out of the intersection.
As for the proof,
$$M \cup (N/M) = M \cup (N \cap \overline M) = (M \cup N) \cap (M \cup \overline M) = M \cup N = N$$
A: First show that if $A\subset C$ and $B\subset C$, then $(A\cup B) \subset C$.
Then we have $M\subset N$ and $(N\setminus M)\subset N$.
To prove $N\subset M\cup (N\setminus M)$, note that $N=(N\cap M) \cup (N\setminus M)\subset M\cup (N\setminus M)$.

Alternatively, we have $M\cup (N\setminus M)=N\cup (M\setminus N)$, and as $M\subset N$, then $M\setminus N=\emptyset$.
A: So clearly if $x\in N\setminus M$ then $x\in N$ and if $x\in M$ then $x\in N$ from the hypothesis $M\subseteq N$ so that
$$M\cup N\setminus M\subseteq N$$
and thus we have only to show the other inclusion. So clearly if $x\in N$ then either $x\in M\cap N=M$ or $x\notin M\cap N=M$ (any event is true or false!) and thus $x\in M$ or $x\in N\setminus M$ so that $x\in M\cup N\setminus M$  and thus the statement follows immediately.
