Intersection with Subset is Subset/Proof 2
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Theorem
- $S \subseteq T \iff S \cap T = S$
Proof
| \(\ds \) | \(\) | \(\ds S \cap T = S\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(\) | \(\ds \paren {x \in S \land x \in T \iff x \in S}\) | Definition of Set Equality | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(\) | \(\ds \paren {x \in S \implies x \in T}\) | Conditional iff Biconditional of Antecedent with Conjunction | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(\) | \(\ds S \subseteq T\) | Definition of Subset |
$\blacksquare$
Sources
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- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): Appendix $\text{A}.2$: Theorem $\text{A}.11$