Set Union Preserves Subsets

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

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

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

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

Proof
Let $A \subseteq B$ and $S \subseteq T$.

Then:

Now we invoke the Constructive Dilemma of propositional logic:
 * $p \implies q, \ r \implies s \vdash p \lor r \implies q \lor s$

applying it as:
 * $\left({x \in A \implies x \in B, \ x \in S \implies x \in T}\right) \implies \left({x \in A \lor x \in S \implies x \in B \lor x \in T}\right)$

The result follows directly from the definition of set union:
 * $\left({x \in A \implies x \in B, \ x \in S \implies x \in T}\right) \implies \left({x \in A \cup S \implies x \in B \cup T}\right)$

and from the definition of subset:
 * $A \subseteq B, \ S \subseteq T \implies A \cup S \subseteq B \cup T$

Proof of Corollary
Let $A \subseteq B$, and let $S$ be any set.

From the main result above, substituting $S$ for $T$:


 * $A \subseteq B, \ S \subseteq S \implies A \cup S \subseteq B \cup S$

From Subset of Itself, $S \subseteq S$ for all sets $S$.

Hence the result:
 * $A \subseteq B \implies A \cup S \subseteq B \cup S$