MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  rcompleq Structured version   Visualization version   GIF version

Theorem rcompleq 4304
Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 4151. (Contributed by RP, 10-Jun-2021.)
Assertion
Ref Expression
rcompleq ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))

Proof of Theorem rcompleq
StepHypRef Expression
1 ancom 460 . . 3 ((𝐴𝐵𝐵𝐴) ↔ (𝐵𝐴𝐴𝐵))
2 sscon34b 4303 . . . . 5 ((𝐵𝐶𝐴𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
32ancoms 458 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
4 sscon34b 4303 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))
53, 4anbi12d 632 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐵𝐴𝐴𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
61, 5bitrid 283 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐴𝐵𝐵𝐴) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
7 eqss 3998 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
8 eqss 3998 . 2 ((𝐶𝐴) = (𝐶𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴)))
96, 7, 83bitr4g 314 1 ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1539  cdif 3947  wss 3950
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2707
This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1542  df-ex 1779  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815  df-rab 3436  df-v 3481  df-dif 3953  df-in 3957  df-ss 3967
This theorem is referenced by:  indifbi  32540  ntrclsfveq1  44078  ntrclsfveq2  44079  ntrclskb  44087  ntrclsk13  44089  ntrclsk4  44090
  Copyright terms: Public domain W3C validator