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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 3178 | . . . . 5 ⊢ (A ∩ (B ∪ 𝐶)) = ((A ∩ B) ∪ (A ∩ 𝐶)) | |
2 | 1 | equncomi 3083 | . . . 4 ⊢ (A ∩ (B ∪ 𝐶)) = ((A ∩ 𝐶) ∪ (A ∩ B)) |
3 | uneq2 3085 | . . . . 5 ⊢ ((A ∩ B) = ∅ → ((A ∩ 𝐶) ∪ (A ∩ B)) = ((A ∩ 𝐶) ∪ ∅)) | |
4 | un0 3245 | . . . . 5 ⊢ ((A ∩ 𝐶) ∪ ∅) = (A ∩ 𝐶) | |
5 | 3, 4 | syl6eq 2085 | . . . 4 ⊢ ((A ∩ B) = ∅ → ((A ∩ 𝐶) ∪ (A ∩ B)) = (A ∩ 𝐶)) |
6 | 2, 5 | syl5eq 2081 | . . 3 ⊢ ((A ∩ B) = ∅ → (A ∩ (B ∪ 𝐶)) = (A ∩ 𝐶)) |
7 | 6 | eqeq1d 2045 | . 2 ⊢ ((A ∩ B) = ∅ → ((A ∩ (B ∪ 𝐶)) = A ↔ (A ∩ 𝐶) = A)) |
8 | df-ss 2925 | . 2 ⊢ (A ⊆ (B ∪ 𝐶) ↔ (A ∩ (B ∪ 𝐶)) = A) | |
9 | df-ss 2925 | . 2 ⊢ (A ⊆ 𝐶 ↔ (A ∩ 𝐶) = A) | |
10 | 7, 8, 9 | 3bitr4g 212 | 1 ⊢ ((A ∩ B) = ∅ → (A ⊆ (B ∪ 𝐶) ↔ A ⊆ 𝐶)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 98 = wceq 1242 ∪ cun 2909 ∩ cin 2910 ⊆ wss 2911 ∅c0 3218 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-in1 544 ax-in2 545 ax-io 629 ax-5 1333 ax-7 1334 ax-gen 1335 ax-ie1 1379 ax-ie2 1380 ax-8 1392 ax-10 1393 ax-11 1394 ax-i12 1395 ax-bndl 1396 ax-4 1397 ax-17 1416 ax-i9 1420 ax-ial 1424 ax-i5r 1425 ax-ext 2019 |
This theorem depends on definitions: df-bi 110 df-tru 1245 df-nf 1347 df-sb 1643 df-clab 2024 df-cleq 2030 df-clel 2033 df-nfc 2164 df-v 2553 df-dif 2914 df-un 2916 df-in 2918 df-ss 2925 df-nul 3219 |
This theorem is referenced by: (None) |
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