Set Union Preserves Subsets/Proof 1

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:
 * $\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \implies \paren {x \in A \lor x \in S \implies x \in B \lor x \in T}$

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

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