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Mirrors > Home > MPE Home > Th. List > Mathboxes > 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 3903. (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 452 | . . 3 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ (𝐵 ⊆ 𝐴 ∧ 𝐴 ⊆ 𝐵)) | |
2 | sscon34b 38843 | . . . . 5 ⊢ ((𝐵 ⊆ 𝐶 ∧ 𝐴 ⊆ 𝐶) → (𝐵 ⊆ 𝐴 ↔ (𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵))) | |
3 | 2 | ancoms 455 | . . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐵 ⊆ 𝐴 ↔ (𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵))) |
4 | sscon34b 38843 | . . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))) | |
5 | 3, 4 | anbi12d 616 | . . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐵 ⊆ 𝐴 ∧ 𝐴 ⊆ 𝐵) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))) |
6 | 1, 5 | syl5bb 272 | . 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))) |
7 | eqss 3767 | . 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴)) | |
8 | eqss 3767 | . 2 ⊢ ((𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵) ↔ ((𝐶 ∖ 𝐴) ⊆ (𝐶 ∖ 𝐵) ∧ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))) | |
9 | 6, 7, 8 | 3bitr4g 303 | 1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 = 𝐵 ↔ (𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 196 ∧ wa 382 = wceq 1631 ∖ cdif 3720 ⊆ wss 3723 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1870 ax-4 1885 ax-5 1991 ax-6 2057 ax-7 2093 ax-9 2154 ax-10 2174 ax-11 2190 ax-12 2203 ax-13 2408 ax-ext 2751 |
This theorem depends on definitions: df-bi 197 df-an 383 df-or 835 df-tru 1634 df-ex 1853 df-nf 1858 df-sb 2050 df-clab 2758 df-cleq 2764 df-clel 2767 df-nfc 2902 df-v 3353 df-dif 3726 df-in 3730 df-ss 3737 |
This theorem is referenced by: ntrclsfveq1 38884 ntrclsfveq2 38885 ntrclskb 38893 ntrclsk13 38895 ntrclsk4 38896 |
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