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| Mirrors > Home > MPE Home > Th. List > disjssun | Structured version Visualization version GIF version | ||
| Description: Subset relation for disjoint classes. (Contributed by NM, 25-Oct-2005.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) |
| Ref | Expression |
|---|---|
| disjssun | ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | uneq2 4112 | . . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = ((𝐴 ∩ 𝐶) ∪ ∅)) | |
| 2 | indi 4234 | . . . . 5 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) | |
| 3 | 2 | equncomi 4110 | . . . 4 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) |
| 4 | un0 4344 | . . . . 5 ⊢ ((𝐴 ∩ 𝐶) ∪ ∅) = (𝐴 ∩ 𝐶) | |
| 5 | 4 | eqcomi 2740 | . . . 4 ⊢ (𝐴 ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∪ ∅) |
| 6 | 1, 3, 5 | 3eqtr4g 2791 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ (𝐵 ∪ 𝐶)) = (𝐴 ∩ 𝐶)) |
| 7 | 6 | eqeq1d 2733 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴 ↔ (𝐴 ∩ 𝐶) = 𝐴)) |
| 8 | dfss2 3920 | . 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴) | |
| 9 | dfss2 3920 | . 2 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐴 ∩ 𝐶) = 𝐴) | |
| 10 | 7, 8, 9 | 3bitr4g 314 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶)) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 ↔ wb 206 = wceq 1541 ∪ cun 3900 ∩ cin 3901 ⊆ wss 3902 ∅c0 4283 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1968 ax-7 2009 ax-8 2113 ax-9 2121 ax-ext 2703 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-3an 1088 df-tru 1544 df-fal 1554 df-ex 1781 df-sb 2068 df-clab 2710 df-cleq 2723 df-clel 2806 df-v 3438 df-dif 3905 df-un 3907 df-in 3909 df-ss 3919 df-nul 4284 |
| This theorem is referenced by: ssfi 9082 hashbclem 14359 alexsubALTlem2 23964 iccntr 24738 reconnlem1 24743 dvne0 25944 |
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