If 'S is a set complement of S, then a set complement of a superset of S' is a subset of S.Just want to verify that above is true just based on the definition of set complement and superset.
set complement and superset
sets
Let $U$ be the universal set. We have $S\subseteq U$. The complementof $S$ is define as $S'=\{x\in U\mid x\notin S\}$. Thus $\forall x\inU,\; x\in S'\text{ iff } x\notin S$.Let $T$ be a superset of $S'$, $T\supseteq S'$, which is true iff $\forall x\inU,\; x\in S'\text{ implies } x\in T$.Then let $T'$ be the complement of $T$, so that again, as in the caseof $S$, $\forall x\in U,\; x\in T'\text{ iff } x\notin T$.Now you want to prove that $T'$ is a subset of $S$, $T'\subseteq S$,which is true iff $\forall x\inU,\; x\in T'\text{ implies } x\in S$.Proof: Consider any element $x\in U$. Taking all our definitions inreverse order, we see that if $x\in T'$ then $x\notin T$. But this inturn that $x\notin S'$ since all elements in $S'$ are in its superset$T$. But by definition of $S'$, all elements not in $S'$ are in $S$.So we have proved that $\forall x\inU,\; x\in T'\text{ implies } x\in S$.Hence  $T'\subseteq S$.