Relative Complement of Relative Complement/Proof 1

Theorem
The relative complement of the relative complement of a set is itself:

If $T \subseteq S$ then:


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

Thus, considered as a mapping on the power set of $S$:
 * $\complement_S: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right): \complement_S \left({T}\right) = S \setminus T$

$\complement_S$ is an involution.

Proof
$\complement_S \left({\complement_S \left({T}\right)}\right) = S \setminus \left({S \setminus T}\right)$ by the definition of relative complement.

Let $t \in T$.

Then $t \notin S \setminus T$ by the definition of set difference.

Since $T \in T$ and $T \subseteq S$, $t \in S$ by the definition of subset.

Thus $t \in \left({S \setminus \left({S \setminus T}\right)}\right)$

Suppose instead that $t \in \left({S \setminus \left({S \setminus T}\right)}\right)$

Then $t \in S$ and $\neg \left({t \in \left({S \setminus T}\right)}\right)$.

Thus $\neg \left({\left({t \in S}\right) \land \neg \left({t \in T}\right)}\right)$.

By Implication Equivalent to Negation of Conjunction with Negative:


 * $t \in S \implies t \in T$

By Modus Ponendo Ponens:


 * $t \in T$