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:
 * $$p \implies q, \ r \implies s \vdash p \or r \implies q \or s$$

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

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