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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 3222 | . . 3 ⊢ (𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶))) | |
2 | elun 3222 | . . . 4 ⊢ (𝑥 ∈ (𝐵 ∪ 𝐶) ↔ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) | |
3 | 2 | orbi2i 752 | . . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))) |
4 | elun 3222 | . . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)) | |
5 | 4 | orbi1i 753 | . . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶)) |
6 | orass 757 | . . . 4 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))) | |
7 | 5, 6 | bitr2i 184 | . . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶)) |
8 | 1, 3, 7 | 3bitrri 206 | . 2 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶))) |
9 | 8 | uneqri 3223 | 1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶)) |
Colors of variables: wff set class |
Syntax hints: ∨ wo 698 = wceq 1332 ∈ wcel 1481 ∪ cun 3074 |
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-io 699 ax-5 1424 ax-7 1425 ax-gen 1426 ax-ie1 1470 ax-ie2 1471 ax-8 1483 ax-10 1484 ax-11 1485 ax-i12 1486 ax-bndl 1487 ax-4 1488 ax-17 1507 ax-i9 1511 ax-ial 1515 ax-i5r 1516 ax-ext 2122 |
This theorem depends on definitions: df-bi 116 df-tru 1335 df-nf 1438 df-sb 1737 df-clab 2127 df-cleq 2133 df-clel 2136 df-nfc 2271 df-v 2691 df-un 3080 |
This theorem is referenced by: un12 3239 un23 3240 un4 3241 qdass 3628 qdassr 3629 rdgisucinc 6290 oasuc 6368 unfidisj 6818 undifdc 6820 djuassen 7090 fzosplitprm1 10042 hashunlem 10582 |
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