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| Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 4151. (Contributed by RP, 10-Jun-2021.) | 
| Ref | Expression | 
|---|---|
| rcompleq | ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 = 𝐵 ↔ (𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵))) | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | ancom 460 | . . 3 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ (𝐵 ⊆ 𝐴 ∧ 𝐴 ⊆ 𝐵)) | |
| 2 | sscon34b 4303 | . . . . 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 4303 | . . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))) | |
| 5 | 3, 4 | anbi12d 632 | . . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐵 ⊆ 𝐴 ∧ 𝐴 ⊆ 𝐵) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))) | 
| 6 | 1, 5 | bitrid 283 | . 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))) | 
| 7 | eqss 3998 | . 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴)) | |
| 8 | eqss 3998 | . 2 ⊢ ((𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))) | |
| 9 | 6, 7, 8 | 3bitr4g 314 | 1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 = 𝐵 ↔ (𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵))) | 
| Colors of variables: wff setvar class | 
| Syntax hints: → wi 4 ↔ wb 206 ∧ wa 395 = wceq 1539 ∖ cdif 3947 ⊆ wss 3950 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1794 ax-4 1808 ax-5 1909 ax-6 1966 ax-7 2006 ax-8 2109 ax-9 2117 ax-ext 2707 | 
| This theorem depends on definitions: df-bi 207 df-an 396 df-3an 1088 df-tru 1542 df-ex 1779 df-sb 2064 df-clab 2714 df-cleq 2728 df-clel 2815 df-rab 3436 df-v 3481 df-dif 3953 df-in 3957 df-ss 3967 | 
| This theorem is referenced by: indifbi 32540 ntrclsfveq1 44078 ntrclsfveq2 44079 ntrclskb 44087 ntrclsk13 44089 ntrclsk4 44090 | 
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