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:
\(\ds x \in A\) | \(\leadsto\) | \(\ds x \in B\) | Definition of Subset | |||||||||||
\(\ds x \in S\) | \(\leadsto\) | \(\ds x \in T\) | 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:
- $A \subseteq B, \ S \subseteq T \implies A \cup S \subseteq B \cup T$
$\blacksquare$