Prove that $A\cup (X \setminus A) = X$ Let $X$ be a set containing $A$.
Proof: 
$y\in A \cup (X \setminus  A) \Rightarrow y\in A$ or $y \in (X \setminus A)$
If $y \in A$, Then $y \in X$ because $A \subset X$. 
If $y \in (X \setminus  A)$, Then $y \in X$ and $y \notin A$. So $y \in X$.
Therefore $y \in A \cup (A \setminus X) \Rightarrow y \in X$.
Now I've proved that every element of $A \cup (A \setminus X)$ is also an element of $X$.
But how to prove vice versa??
 A: The reverse implication is simple enough.
If $y \in X$, then either $y \in A$ or $y \not \in A$, since $A$ is a subset of $X$.
If $y \in A$, then $y \in X$ since $A \subseteq X$ is given.
If $y \not \in A$, since $y \in X$ is also given, then $y \in X - A$.
Therefore, in either case, $y \in A \cup (X-A)$, the desired conclusion.
A: Let $A\subseteq X$
$\Leftarrow:$
Show
$$X\subseteq A\cup(X-A)$$
Let $\phi_1,\phi_2$ be some formula
write $A=\{a:\phi_1(a)\}, X=\{x:\phi_2(x)\}$
Since $A\subseteq X$ we have for any $t$ that $\phi_1(t)$ implies $\phi_2(t)$
Note that $$A\cup(X-A)=\{t:\phi_1(t)\vee(\phi_2(t)\wedge\neg\phi_1(t))\}$$
To show $X$ is a subset of $A\cup(X-A)$ is same as to show that
$$\forall t,\phi_2(t)\rightarrow\phi_1(t)\vee(\phi_2(t)\wedge\neg\phi_1(t))$$
Which is clearly a tautology
Hence proved
$\Rightarrow:$
Similarly, for another direction we also need the condition $A\subseteq X$, just consider
$$\forall t,\big((\phi_1(t)\rightarrow\phi_2(t))\wedge(\phi_1(t)\vee(\phi_2(t)\wedge\neg\phi_1(t)))\big)\rightarrow\phi_2(t)$$
Not hard to see, it's also a tautology
Which proved $A\cup(X-A)\subseteq X\tag*{$\square$}$
A: The inverse is really trivial.
Proposition:  For any sets $A, B$ than $A \subset A\cup B$.
Pf: If $x \in A$ then ($x\in A$ or $x \in B$).  So $x\in A\cup B$.
So $A\subset A \cup (A\setminus X)$
