Theorem cleq1 13722

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

Assertion

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

Distinct variable groups:   R, r   S, r

Allowed substitution hint:   ph( r)

Proof of Theorem cleq1

Step	Hyp	Ref	Expression
1		cleq1lem 13721	. . 3 |- ( R = S -> ( ( R C_ r /\ ph ) <-> ( S C_ r /\ ph ) ) )
2	1	abbidv 2741	. 2 |- ( R = S -> { r | ( R C_ r /\ ph ) } = { r | ( S C_ r /\ ph ) } )
3	2	inteqd 4480	1 |- ( R = S -> |^| { r | ( R C_ r /\ ph ) } = |^| { r | ( S C_ r /\ ph ) } )

Syntax hints:   -> wi 4   /\ wa 384   = wceq 1483   {cab 2608   C_ wss 3574   |^|cint 4475

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ral 2917  df-in 3581  df-ss 3588  df-int 4476

This theorem is referenced by:  trcleq1  13728