Union is Empty iff Sets are Empty/Proof 1
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Theorem
- $S \cup T = \O \iff S = \O \land T = \O$
Proof
\(\ds S \cup T = \varnothing\) | \(\iff\) | \(\ds \neg \exists x: x \in \left ({S \cup T}\right)\) | Definition of Empty Set | |||||||||||
\(\ds \) | \(\iff\) | \(\ds \forall x: \neg \left ({x \in \left ({S \cup T}\right)}\right)\) | De Morgan's Laws (Predicate Logic) | |||||||||||
\(\ds \) | \(\iff\) | \(\ds \forall x: \neg \left ({x \in S \lor x \in T}\right)\) | Definition of Set Union | |||||||||||
\(\ds \) | \(\iff\) | \(\ds \forall x: x \notin S \land x \notin T\) | De Morgan's Laws: Conjunction of Negations | |||||||||||
\(\ds \) | \(\iff\) | \(\ds S = \varnothing \land T = \varnothing\) | Definition of Empty Set |
$\blacksquare$