Set Union Preserves Subsets/Proof 1

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

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

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