Set Union Preserves Subsets/Corollary/Proof 2

Theorem
Let $A, B, S$ be sets.

Then:
 * $A \subseteq B \implies A \cup S \subseteq B \cup S$
 * $A \subseteq B \implies S \cup A \subseteq S \cup B$

Proof
Let $A$, $B$, and $S$ be sets.

Let $A \subseteq B$.

Let $x \in A \cup S$.

By the definition of union:
 * $x \in A$ or $x \in S$.

Suppose $x \in A$.

Then by the definition of subset, $x \in B$.

Thus by the definition of union, $x \in B \cup S$.

Suppose instead that $x \in S$.

Then by the definition of union, $x \in B \cup S$.

Thus for all $x \in A \cup S$, $x \in B \cup S$

The second result follows from Union is Commutative.