Intersection with Subset is Subset/Proof 2

From ProofWiki
Jump to navigation Jump to search

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