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Theorem rcompleq 4254
Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 4106. (Contributed by RP, 10-Jun-2021.)
Assertion
Ref Expression
rcompleq ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))

Proof of Theorem rcompleq
StepHypRef Expression
1 ancom 461 . . 3 ((𝐴𝐵𝐵𝐴) ↔ (𝐵𝐴𝐴𝐵))
2 sscon34b 4253 . . . . 5 ((𝐵𝐶𝐴𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
32ancoms 459 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
4 sscon34b 4253 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))
53, 4anbi12d 631 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐵𝐴𝐴𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
61, 5bitrid 282 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐴𝐵𝐵𝐴) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
7 eqss 3958 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
8 eqss 3958 . 2 ((𝐶𝐴) = (𝐶𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴)))
96, 7, 83bitr4g 313 1 ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 396   = wceq 1541  cdif 3906  wss 3909
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2707
This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1544  df-ex 1782  df-sb 2068  df-clab 2714  df-cleq 2728  df-clel 2814  df-rab 3407  df-v 3446  df-dif 3912  df-in 3916  df-ss 3926
This theorem is referenced by:  indifbi  31345  ntrclsfveq1  42312  ntrclsfveq2  42313  ntrclskb  42321  ntrclsk13  42323  ntrclsk4  42324
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