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Mirrors > Home > MPE Home > Th. List > rcompleq | Structured version Visualization version GIF version |
Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 4162. (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 4310 | . . . . 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 4310 | . . . 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 4011 | . 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴)) | |
8 | eqss 4011 | . 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 1537 ∖ cdif 3960 ⊆ wss 3963 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1908 ax-6 1965 ax-7 2005 ax-8 2108 ax-9 2116 ax-ext 2706 |
This theorem depends on definitions: df-bi 207 df-an 396 df-3an 1088 df-tru 1540 df-ex 1777 df-sb 2063 df-clab 2713 df-cleq 2727 df-clel 2814 df-rab 3434 df-v 3480 df-dif 3966 df-in 3970 df-ss 3980 |
This theorem is referenced by: indifbi 32548 ntrclsfveq1 44050 ntrclsfveq2 44051 ntrclskb 44059 ntrclsk13 44061 ntrclsk4 44062 |
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