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Mirrors > Home > ILE Home > Th. List > unass | GIF version |
Description: Associative law for union of classes. Exercise 8 of [TakeutiZaring] p. 17. (Contributed by NM, 3-May-1994.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) |
Ref | Expression |
---|---|
unass | ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | elun 3114 | . . 3 ⊢ (𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶))) | |
2 | elun 3114 | . . . 4 ⊢ (𝑥 ∈ (𝐵 ∪ 𝐶) ↔ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) | |
3 | 2 | orbi2i 712 | . . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))) |
4 | elun 3114 | . . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)) | |
5 | 4 | orbi1i 713 | . . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶)) |
6 | orass 717 | . . . 4 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))) | |
7 | 5, 6 | bitr2i 183 | . . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶)) |
8 | 1, 3, 7 | 3bitrri 205 | . 2 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶))) |
9 | 8 | uneqri 3115 | 1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶)) |
Colors of variables: wff set class |
Syntax hints: ∨ wo 662 = wceq 1285 ∈ wcel 1434 ∪ cun 2972 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-io 663 ax-5 1377 ax-7 1378 ax-gen 1379 ax-ie1 1423 ax-ie2 1424 ax-8 1436 ax-10 1437 ax-11 1438 ax-i12 1439 ax-bndl 1440 ax-4 1441 ax-17 1460 ax-i9 1464 ax-ial 1468 ax-i5r 1469 ax-ext 2064 |
This theorem depends on definitions: df-bi 115 df-tru 1288 df-nf 1391 df-sb 1687 df-clab 2069 df-cleq 2075 df-clel 2078 df-nfc 2209 df-v 2604 df-un 2978 |
This theorem is referenced by: un12 3131 un23 3132 un4 3133 qdass 3491 qdassr 3492 rdgisucinc 6028 oasuc 6102 unfidisj 6432 undiffi 6433 fzosplitprm1 9309 sizeunlem 9817 |
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