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Theorem compleq 4123
 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 4122 . . 3 (𝐴𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2 complss 4122 . . 3 (𝐵𝐴 ↔ (V ∖ 𝐴) ⊆ (V ∖ 𝐵))
31, 2anbi12ci 629 . 2 ((𝐴𝐵𝐵𝐴) ↔ ((V ∖ 𝐴) ⊆ (V ∖ 𝐵) ∧ (V ∖ 𝐵) ⊆ (V ∖ 𝐴)))
4 eqss 3981 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
5 eqss 3981 . 2 ((V ∖ 𝐴) = (V ∖ 𝐵) ↔ ((V ∖ 𝐴) ⊆ (V ∖ 𝐵) ∧ (V ∖ 𝐵) ⊆ (V ∖ 𝐴)))
63, 4, 53bitr4i 305 1 (𝐴 = 𝐵 ↔ (V ∖ 𝐴) = (V ∖ 𝐵))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 208   ∧ wa 398   = wceq 1533  Vcvv 3494   ∖ cdif 3932   ⊆ wss 3935 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793 This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-v 3496  df-dif 3938  df-in 3942  df-ss 3951 This theorem is referenced by: (None)
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