Set Union Preserves Subsets/Proof 1

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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:

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

$\blacksquare$