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Theorem rcompleq 37797
Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 3730. (Contributed by RP, 10-Jun-2021.)
Assertion
Ref Expression
rcompleq ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))

Proof of Theorem rcompleq
StepHypRef Expression
1 ancom 466 . . 3 ((𝐴𝐵𝐵𝐴) ↔ (𝐵𝐴𝐴𝐵))
2 sscon34b 37796 . . . . 5 ((𝐵𝐶𝐴𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
32ancoms 469 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
4 sscon34b 37796 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))
53, 4anbi12d 746 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐵𝐴𝐴𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
61, 5syl5bb 272 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐴𝐵𝐵𝐴) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
7 eqss 3598 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
8 eqss 3598 . 2 ((𝐶𝐴) = (𝐶𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴)))
96, 7, 83bitr4g 303 1 ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wa 384   = wceq 1480  cdif 3552  wss 3555
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-v 3188  df-dif 3558  df-in 3562  df-ss 3569
This theorem is referenced by:  ntrclsfveq1  37837  ntrclsfveq2  37838  ntrclskb  37846  ntrclsk13  37848  ntrclsk4  37849
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