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Theorem rcompleq 4297
Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 4147. (Contributed by RP, 10-Jun-2021.)
Assertion
Ref Expression
rcompleq ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))

Proof of Theorem rcompleq
StepHypRef Expression
1 ancom 459 . . 3 ((𝐴𝐵𝐵𝐴) ↔ (𝐵𝐴𝐴𝐵))
2 sscon34b 4296 . . . . 5 ((𝐵𝐶𝐴𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
32ancoms 457 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
4 sscon34b 4296 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))
53, 4anbi12d 630 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐵𝐴𝐴𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
61, 5bitrid 282 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐴𝐵𝐵𝐴) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
7 eqss 3995 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
8 eqss 3995 . 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 394   = wceq 1534  cdif 3944  wss 3947
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2697
This theorem depends on definitions:  df-bi 206  df-an 395  df-3an 1086  df-tru 1537  df-ex 1775  df-sb 2061  df-clab 2704  df-cleq 2718  df-clel 2803  df-rab 3420  df-v 3464  df-dif 3950  df-in 3954  df-ss 3964
This theorem is referenced by:  indifbi  32447  ntrclsfveq1  43727  ntrclsfveq2  43728  ntrclskb  43736  ntrclsk13  43738  ntrclsk4  43739
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