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Metamath Proof Explorer

Theorem petlemi

Description: If you can prove disjointness (e.g. disjALTV0 , disjALTVid , disjALTVidres , disjALTVxrnidres , search for theorems containing the ' |- Disj ' string), or the same with converse function (cf. dfdisjALTV ), then disjointness, and equivalence of cosets, both on their natural domain, are equivalent. (Contributed by Peter Mazsa, 18-Sep-2021)

		Ref	Expression
	Hypothesis	petlemi.1	\|- Disj R
	Assertion	petlemi	\|- ( ( Disj R /\ ( dom R /. R ) = A ) <-> ( EqvRel ,~ R /\ ( dom ,~ R /. ,~ R ) = A ) )

Proof

Step	Hyp	Ref	Expression
1		petlemi.1	\|- Disj R
2	1	a1i	\|- ( ( EqvRel ,~ R /\ ( dom ,~ R /. ,~ R ) = A ) -> Disj R )
3	2	petlem	\|- ( ( Disj R /\ ( dom R /. R ) = A ) <-> ( EqvRel ,~ R /\ ( dom ,~ R /. ,~ R ) = A ) )