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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 3957 | . . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = ((𝐴 ∩ 𝐶) ∪ ∅)) | |
2 | indi 4072 | . . . . 5 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) | |
3 | 2 | equncomi 3955 | . . . 4 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) |
4 | un0 4161 | . . . . 5 ⊢ ((𝐴 ∩ 𝐶) ∪ ∅) = (𝐴 ∩ 𝐶) | |
5 | 4 | eqcomi 2806 | . . . 4 ⊢ (𝐴 ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∪ ∅) |
6 | 1, 3, 5 | 3eqtr4g 2856 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ (𝐵 ∪ 𝐶)) = (𝐴 ∩ 𝐶)) |
7 | 6 | eqeq1d 2799 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴 ↔ (𝐴 ∩ 𝐶) = 𝐴)) |
8 | df-ss 3781 | . 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴) | |
9 | df-ss 3781 | . 2 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐴 ∩ 𝐶) = 𝐴) | |
10 | 7, 8, 9 | 3bitr4g 306 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶)) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 198 = wceq 1653 ∪ cun 3765 ∩ cin 3766 ⊆ wss 3767 ∅c0 4113 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1891 ax-4 1905 ax-5 2006 ax-6 2072 ax-7 2107 ax-9 2166 ax-10 2185 ax-11 2200 ax-12 2213 ax-ext 2775 |
This theorem depends on definitions: df-bi 199 df-an 386 df-or 875 df-tru 1657 df-ex 1876 df-nf 1880 df-sb 2065 df-clab 2784 df-cleq 2790 df-clel 2793 df-nfc 2928 df-v 3385 df-dif 3770 df-un 3772 df-in 3774 df-ss 3781 df-nul 4114 |
This theorem is referenced by: hashbclem 13481 alexsubALTlem2 22177 iccntr 22949 reconnlem1 22954 dvne0 24112 |
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