Theorem compleq 4106

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 4105	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2		complss 4105	. . 3 ⊢ (𝐵 ⊆ 𝐴 ↔ (V ∖ 𝐴) ⊆ (V ∖ 𝐵))
3	1, 2	anbi12ci 641	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ ((V ∖ 𝐴) ⊆ (V ∖ 𝐵) ∧ (V ∖ 𝐵) ⊆ (V ∖ 𝐴)))
4		eqss 3953	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 3953	. 2 ⊢ ((V ∖ 𝐴) = (V ∖ 𝐵) ↔ ((V ∖ 𝐴) ⊆ (V ∖ 𝐵) ∧ (V ∖ 𝐵) ⊆ (V ∖ 𝐴)))
6	3, 4, 5	3bitr4i 306	1 ⊢ (𝐴 = 𝐵 ↔ (V ∖ 𝐴) = (V ∖ 𝐵))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570  Vcvv 3457   ∖ cdif 3903   ⊆ wss 3906

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-v 3459  df-dif 3909  df-ss 3923

This theorem is used by: (None)