Theorem cleq1

Description: Equality of relations implies equality of closures. (Contributed by RP, 9-May-2020)

	Ref	Expression
Assertion	cleq1	|- ( R = S -> |^| { r | ( R C_ r /\ ph ) } = |^| { r | ( S C_ r /\ ph ) } )

Proof

Step	Hyp	Ref	Expression
1		cleq1lem	|- ( R = S -> ( ( R C_ r /\ ph ) <-> ( S C_ r /\ ph ) ) )
2	1	abbidv	|- ( R = S -> { r | ( R C_ r /\ ph ) } = { r | ( S C_ r /\ ph ) } )
3	2	inteqd	|- ( R = S -> |^| { r | ( R C_ r /\ ph ) } = |^| { r | ( S C_ r /\ ph ) } )