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| Mirrors > Home > MPE Home > Th. List > sscon34b | Structured version Visualization version GIF version | ||
| Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss 4113. (Contributed by RP, 3-Jun-2021.) |
| Ref | Expression |
|---|---|
| sscon34b | ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | sscon 4105 | . 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)) | |
| 2 | sscon 4105 | . . 3 ⊢ ((𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴) → (𝐶 ∖ (𝐶 ∖ 𝐴)) ⊆ (𝐶 ∖ (𝐶 ∖ 𝐵))) | |
| 3 | dfss4 4230 | . . . . 5 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴) | |
| 4 | 3 | birani 508 | . . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴) |
| 5 | dfss4 4230 | . . . . 5 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵) | |
| 6 | 5 | bilani 509 | . . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵) |
| 7 | 4, 6 | sseq12d 3978 | . . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ (𝐶 ∖ 𝐴)) ⊆ (𝐶 ∖ (𝐶 ∖ 𝐵)) ↔ 𝐴 ⊆ 𝐵)) |
| 8 | 2, 7 | imbitrid 247 | . 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴) → 𝐴 ⊆ 𝐵)) |
| 9 | 1, 8 | impbid2 229 | 1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 ↔ wb 209 ∧ wa 400 = wceq 1567 ∖ cdif 3910 ⊆ wss 3913 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1822 ax-4 1836 ax-5 1937 ax-6 1994 ax-7 2035 ax-8 2151 ax-9 2159 ax-ext 2741 |
| This theorem depends on definitions: df-bi 210 df-an 401 df-3an 1103 df-tru 1570 df-ex 1807 df-sb 2098 df-clab 2748 df-cleq 2761 df-clel 2844 df-rab 3424 df-v 3465 df-dif 3916 df-in 3920 df-ss 3930 |
| This theorem is referenced by: rcompleq 4266 ntrclsss 44674 ntrclsiso 44678 ntrclsk2 44679 ntrclsk3 44681 |
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