# 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:

 $\displaystyle x \in A$ $\leadsto$ $\displaystyle x \in B$ Definition of Subset $\displaystyle x \in S$ $\leadsto$ $\displaystyle x \in T$ Definition of Subset

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:

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

$\blacksquare$