Union with Relative Complement

Theorem
The union of a set $T$ and its relative complement in $S$ is the set $S$:


 * $\complement_S \left({T}\right) \cup T = S$

Proof
From the definition of relative complement, we have that $T \subseteq S$.

From Union with Superset is Superset‎, we have that $T \subseteq S \iff S \cup T = S$, from which the result follows.

$\complement_S \left({T}\right) \cup T\subseteq S$
By the definition of relative complement, we have that $\complement_S \left({T}\right)\subseteq S$ and $T \subseteq S$.

Hence by Union Smallest, $\complement_S \left({T}\right) \cup T\subseteq S$.

$S\subseteq \complement_S \left({T}\right) \cup T$
Let $x \in S$.

By the Law of the Excluded Middle, one of the following two applies:


 * $(1): \quad x \in T$
 * $(2): \quad x \notin T$

If $(2)$, then by definition of relative complement, $x \in S \setminus T = \complement_S \left({T}\right)$.

So $x \in T \lor x \in \complement_S \left({T}\right)$.

By definition of set union, $x\in \complement_S \left({T}\right) \cup T$.

Thus $x \in S \implies x \in \complement_S \left({T}\right) \cup T$.

By definition of subset it follows that $S \subseteq \complement_S \left({T}\right) \cup T$.

Final resolution
From:
 * $\complement_S \left({T}\right) \cup T\subseteq S$

and:
 * $S\subseteq \complement_S \left({T}\right) \cup T$

it follows from Equality of Sets that $S = \complement_S \left({T}\right) \cup T$.