Relative Complement of Relative Complement/Proof 1

Proof
By the definition of relative complement:
 * $\complement_S \left({\complement_S \left({T}\right)}\right) = S \setminus \left({S \setminus T}\right)$

Let $t \in T$.

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

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

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:


 * $t \in T$