Theorem rcompleq 4254

Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 4102. (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 466	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ (𝐵 ⊆ 𝐴 ∧ 𝐴 ⊆ 𝐵))
2		sscon34b 4253	. . . . 5 ⊢ ((𝐵 ⊆ 𝐶 ∧ 𝐴 ⊆ 𝐶) → (𝐵 ⊆ 𝐴 ↔ (𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵)))
3	2	ancoms 464	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐵 ⊆ 𝐴 ↔ (𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵)))
4		sscon34b 4253	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))
5	3, 4	anbi12d 644	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐵 ⊆ 𝐴 ∧ 𝐴 ⊆ 𝐵) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))))
6	1, 5	bitrid 286	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))))
7		eqss 3949	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
8		eqss 3949	. 2 ⊢ ((𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))
9	6, 7, 8	3bitr4g 317	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 = 𝐵 ↔ (𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∖ cdif 3899   ⊆ wss 3902

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 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3455  df-dif 3905  df-in 3909  df-ss 3919

This theorem is used by:  indifbi  32981  ntrclsfveq1  44887  ntrclsfveq2  44888  ntrclskb  44896  ntrclsk13  44898  ntrclsk4  44899