Set Complement inverts Subsets/Proof 1
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Theorem
- $S \subseteq T \iff \map \complement T \subseteq \map \complement S$
Proof
| \(\ds S\) | \(\subseteq\) | \(\ds T\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds S \cap T\) | \(=\) | \(\ds S\) | Intersection with Subset is Subset | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map \complement {S \cap T}\) | \(=\) | \(\ds \map \complement S\) | Complement of Complement | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map \complement S \cup \map \complement T\) | \(=\) | \(\ds \map \complement S\) | De Morgan's Laws: Complement of Intersection | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map \complement T\) | \(\subseteq\) | \(\ds \map \complement S\) | Union with Superset is Superset |
$\blacksquare$