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| Description: Distributive law for union over intersection. Exercise 11 of [TakeutiZaring] p. 17. (Contributed by NM, 30-Sep-2002.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) | 
| Ref | Expression | 
|---|---|
| undi | ⊢ (𝐴 ∪ (𝐵 ∩ 𝐶)) = ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)) | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | elin 3966 | . . . 4 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)) | |
| 2 | 1 | orbi2i 912 | . . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))) | 
| 3 | ordi 1007 | . . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶))) | |
| 4 | elin 3966 | . . . 4 ⊢ (𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝑥 ∈ (𝐴 ∪ 𝐶))) | |
| 5 | elun 4152 | . . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)) | |
| 6 | elun 4152 | . . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)) | |
| 7 | 5, 6 | anbi12i 628 | . . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝑥 ∈ (𝐴 ∪ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶))) | 
| 8 | 4, 7 | bitr2i 276 | . . 3 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)) ↔ 𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶))) | 
| 9 | 2, 3, 8 | 3bitri 297 | . 2 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ 𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶))) | 
| 10 | 9 | uneqri 4155 | 1 ⊢ (𝐴 ∪ (𝐵 ∩ 𝐶)) = ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)) | 
| Colors of variables: wff setvar class | 
| Syntax hints: ∧ wa 395 ∨ wo 847 = wceq 1539 ∈ wcel 2107 ∪ cun 3948 ∩ cin 3949 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1794 ax-4 1808 ax-5 1909 ax-6 1966 ax-7 2006 ax-8 2109 ax-9 2117 ax-ext 2707 | 
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-tru 1542 df-ex 1779 df-sb 2064 df-clab 2714 df-cleq 2728 df-clel 2815 df-v 3481 df-un 3955 df-in 3957 | 
| This theorem is referenced by: undir 4286 dfif4 4540 dfif5 4541 | 
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