Theorem compleq 4132

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 4131	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2		complss 4131	. . 3 ⊢ (𝐵 ⊆ 𝐴 ↔ (V ∖ 𝐴) ⊆ (V ∖ 𝐵))
3	1, 2	anbi12ci 629	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ ((V ∖ 𝐴) ⊆ (V ∖ 𝐵) ∧ (V ∖ 𝐵) ⊆ (V ∖ 𝐴)))
4		eqss 3979	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 3979	. 2 ⊢ ((V ∖ 𝐴) = (V ∖ 𝐵) ↔ ((V ∖ 𝐴) ⊆ (V ∖ 𝐵) ∧ (V ∖ 𝐵) ⊆ (V ∖ 𝐴)))
6	3, 4, 5	3bitr4i 303	1 ⊢ (𝐴 = 𝐵 ↔ (V ∖ 𝐴) = (V ∖ 𝐵))

Syntax hints:   ↔ wb 206   ∧ wa 395   = wceq 1540  Vcvv 3464   ∖ cdif 3928   ⊆ wss 3931

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2715  df-cleq 2728  df-clel 2810  df-v 3466  df-dif 3934  df-ss 3948

This theorem is referenced by: (None)