Theorem compleq 4078

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

Step	Hyp	Ref	Expression
1		complss 4077	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2		complss 4077	. . 3 ⊢ (𝐵 ⊆ 𝐴 ↔ (V ∖ 𝐴) ⊆ (V ∖ 𝐵))
3	1, 2	anbi12ci 627	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ ((V ∖ 𝐴) ⊆ (V ∖ 𝐵) ∧ (V ∖ 𝐵) ⊆ (V ∖ 𝐴)))
4		eqss 3932	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 3932	. 2 ⊢ ((V ∖ 𝐴) = (V ∖ 𝐵) ↔ ((V ∖ 𝐴) ⊆ (V ∖ 𝐵) ∧ (V ∖ 𝐵) ⊆ (V ∖ 𝐴)))
6	3, 4, 5	3bitr4i 302	1 ⊢ (𝐴 = 𝐵 ↔ (V ∖ 𝐴) = (V ∖ 𝐵))

Syntax hints:   ↔ wb 205   ∧ wa 395   = wceq 1539  Vcvv 3422   ∖ cdif 3880   ⊆ wss 3883

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-v 3424  df-dif 3886  df-in 3890  df-ss 3900

This theorem is referenced by: (None)