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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 4154 | . . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = ((𝐴 ∩ 𝐶) ∪ ∅)) | |
2 | indi 4272 | . . . . 5 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) | |
3 | 2 | equncomi 4152 | . . . 4 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) |
4 | un0 4392 | . . . . 5 ⊢ ((𝐴 ∩ 𝐶) ∪ ∅) = (𝐴 ∩ 𝐶) | |
5 | 4 | eqcomi 2734 | . . . 4 ⊢ (𝐴 ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∪ ∅) |
6 | 1, 3, 5 | 3eqtr4g 2790 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ (𝐵 ∪ 𝐶)) = (𝐴 ∩ 𝐶)) |
7 | 6 | eqeq1d 2727 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴 ↔ (𝐴 ∩ 𝐶) = 𝐴)) |
8 | dfss2 3962 | . 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴) | |
9 | dfss2 3962 | . 2 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐴 ∩ 𝐶) = 𝐴) | |
10 | 7, 8, 9 | 3bitr4g 313 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶)) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 205 = wceq 1533 ∪ cun 3942 ∩ cin 3943 ⊆ wss 3944 ∅c0 4322 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1789 ax-4 1803 ax-5 1905 ax-6 1963 ax-7 2003 ax-8 2100 ax-9 2108 ax-ext 2696 |
This theorem depends on definitions: df-bi 206 df-an 395 df-or 846 df-3an 1086 df-tru 1536 df-fal 1546 df-ex 1774 df-sb 2060 df-clab 2703 df-cleq 2717 df-clel 2802 df-v 3463 df-dif 3947 df-un 3949 df-in 3951 df-ss 3961 df-nul 4323 |
This theorem is referenced by: ssfi 9198 hashbclem 14447 alexsubALTlem2 23996 iccntr 24781 reconnlem1 24786 dvne0 25988 |
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