Theorem rcompleq 38318

Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 3752. (Contributed by RP, 10-Jun-2021.)

Assertion

Ref	Expression
rcompleq	|- ( ( A C_ C /\ B C_ C ) -> ( A = B <-> ( C \ A ) = ( C \ B ) ) )

Proof of Theorem rcompleq

Step	Hyp	Ref	Expression
1		ancom 466	. . 3 |- ( ( A C_ B /\ B C_ A ) <-> ( B C_ A /\ A C_ B ) )
2		sscon34b 38317	. . . . 5 |- ( ( B C_ C /\ A C_ C ) -> ( B C_ A <-> ( C \ A ) C_ ( C \ B ) ) )
3	2	ancoms 469	. . . 4 |- ( ( A C_ C /\ B C_ C ) -> ( B C_ A <-> ( C \ A ) C_ ( C \ B ) ) )
4		sscon34b 38317	. . . 4 |- ( ( A C_ C /\ B C_ C ) -> ( A C_ B <-> ( C \ B ) C_ ( C \ A ) ) )
5	3, 4	anbi12d 747	. . 3 |- ( ( A C_ C /\ B C_ C ) -> ( ( B C_ A /\ A C_ B ) <-> ( ( C \ A ) C_ ( C \ B ) /\ ( C \ B ) C_ ( C \ A ) ) ) )
6	1, 5	syl5bb 272	. 2 |- ( ( A C_ C /\ B C_ C ) -> ( ( A C_ B /\ B C_ A ) <-> ( ( C \ A ) C_ ( C \ B ) /\ ( C \ B ) C_ ( C \ A ) ) ) )
7		eqss 3618	. 2 |- ( A = B <-> ( A C_ B /\ B C_ A ) )
8		eqss 3618	. 2 |- ( ( C \ A ) = ( C \ B ) <-> ( ( C \ A ) C_ ( C \ B ) /\ ( C \ B ) C_ ( C \ A ) ) )
9	6, 7, 8	3bitr4g 303	1 |- ( ( A C_ C /\ B C_ C ) -> ( A = B <-> ( C \ A ) = ( C \ B ) ) )

Syntax hints:   -> wi 4   <-> wb 196   /\ wa 384   = wceq 1483   \ cdif 3571   C_ wss 3574

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-v 3202  df-dif 3577  df-in 3581  df-ss 3588

This theorem is referenced by:  ntrclsfveq1  38358  ntrclsfveq2  38359  ntrclskb  38367  ntrclsk13  38369  ntrclsk4  38370