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Mirrors > Home > ILE Home > Th. List > disjssun | 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 | indi 3293 | . . . . 5 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) | |
2 | 1 | equncomi 3192 | . . . 4 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) |
3 | uneq2 3194 | . . . . 5 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = ((𝐴 ∩ 𝐶) ∪ ∅)) | |
4 | un0 3366 | . . . . 5 ⊢ ((𝐴 ∩ 𝐶) ∪ ∅) = (𝐴 ∩ 𝐶) | |
5 | 3, 4 | syl6eq 2166 | . . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = (𝐴 ∩ 𝐶)) |
6 | 2, 5 | syl5eq 2162 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ (𝐵 ∪ 𝐶)) = (𝐴 ∩ 𝐶)) |
7 | 6 | eqeq1d 2126 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴 ↔ (𝐴 ∩ 𝐶) = 𝐴)) |
8 | df-ss 3054 | . 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴) | |
9 | df-ss 3054 | . 2 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐴 ∩ 𝐶) = 𝐴) | |
10 | 7, 8, 9 | 3bitr4g 222 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 104 = wceq 1316 ∪ cun 3039 ∩ cin 3040 ⊆ wss 3041 ∅c0 3333 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 588 ax-in2 589 ax-io 683 ax-5 1408 ax-7 1409 ax-gen 1410 ax-ie1 1454 ax-ie2 1455 ax-8 1467 ax-10 1468 ax-11 1469 ax-i12 1470 ax-bndl 1471 ax-4 1472 ax-17 1491 ax-i9 1495 ax-ial 1499 ax-i5r 1500 ax-ext 2099 |
This theorem depends on definitions: df-bi 116 df-tru 1319 df-nf 1422 df-sb 1721 df-clab 2104 df-cleq 2110 df-clel 2113 df-nfc 2247 df-v 2662 df-dif 3043 df-un 3045 df-in 3047 df-ss 3054 df-nul 3334 |
This theorem is referenced by: (None) |
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