Theorem rcompleq 4226

Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 4078. (Contributed by RP, 10-Jun-2021.)

Assertion

Ref	Expression
rcompleq	⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 = 𝐵 ↔ (𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵)))

Proof of Theorem rcompleq

Step	Hyp	Ref	Expression
1		ancom 460	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ (𝐵 ⊆ 𝐴 ∧ 𝐴 ⊆ 𝐵))
2		sscon34b 4225	. . . . 5 ⊢ ((𝐵 ⊆ 𝐶 ∧ 𝐴 ⊆ 𝐶) → (𝐵 ⊆ 𝐴 ↔ (𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵)))
3	2	ancoms 458	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐵 ⊆ 𝐴 ↔ (𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵)))
4		sscon34b 4225	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))
5	3, 4	anbi12d 630	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐵 ⊆ 𝐴 ∧ 𝐴 ⊆ 𝐵) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))))
6	1, 5	syl5bb 282	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))))
7		eqss 3932	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
8		eqss 3932	. 2 ⊢ ((𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))
9	6, 7, 8	3bitr4g 313	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 = 𝐵 ↔ (𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1539   ∖ 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-rab 3072  df-v 3424  df-dif 3886  df-in 3890  df-ss 3900

This theorem is referenced by:  indifbi  30769  ntrclsfveq1  41559  ntrclsfveq2  41560  ntrclskb  41568  ntrclsk13  41570  ntrclsk4  41571