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Theorem compleq 3730
Description: Two classes are equal if and only if their complements are equal. (Contributed by BJ, 19-Mar-2021.)
Assertion
Ref Expression
compleq (𝐴 = 𝐵 ↔ (V ∖ 𝐴) = (V ∖ 𝐵))

Proof of Theorem compleq
StepHypRef Expression
1 complss 3729 . . 3 (𝐴𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2 complss 3729 . . 3 (𝐵𝐴 ↔ (V ∖ 𝐴) ⊆ (V ∖ 𝐵))
31, 2anbi12ci 733 . 2 ((𝐴𝐵𝐵𝐴) ↔ ((V ∖ 𝐴) ⊆ (V ∖ 𝐵) ∧ (V ∖ 𝐵) ⊆ (V ∖ 𝐴)))
4 eqss 3598 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
5 eqss 3598 . 2 ((V ∖ 𝐴) = (V ∖ 𝐵) ↔ ((V ∖ 𝐴) ⊆ (V ∖ 𝐵) ∧ (V ∖ 𝐵) ⊆ (V ∖ 𝐴)))
63, 4, 53bitr4i 292 1 (𝐴 = 𝐵 ↔ (V ∖ 𝐴) = (V ∖ 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wb 196  wa 384   = wceq 1480  Vcvv 3186  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: (None)
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