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Mirrors > Home > ILE Home > Th. List > undi | GIF version |
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 3120 | . . . 4 ⊢ (x ∈ (B ∩ 𝐶) ↔ (x ∈ B ∧ x ∈ 𝐶)) | |
2 | 1 | orbi2i 678 | . . 3 ⊢ ((x ∈ A ∨ x ∈ (B ∩ 𝐶)) ↔ (x ∈ A ∨ (x ∈ B ∧ x ∈ 𝐶))) |
3 | ordi 728 | . . 3 ⊢ ((x ∈ A ∨ (x ∈ B ∧ x ∈ 𝐶)) ↔ ((x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ 𝐶))) | |
4 | elin 3120 | . . . 4 ⊢ (x ∈ ((A ∪ B) ∩ (A ∪ 𝐶)) ↔ (x ∈ (A ∪ B) ∧ x ∈ (A ∪ 𝐶))) | |
5 | elun 3078 | . . . . 5 ⊢ (x ∈ (A ∪ B) ↔ (x ∈ A ∨ x ∈ B)) | |
6 | elun 3078 | . . . . 5 ⊢ (x ∈ (A ∪ 𝐶) ↔ (x ∈ A ∨ x ∈ 𝐶)) | |
7 | 5, 6 | anbi12i 433 | . . . 4 ⊢ ((x ∈ (A ∪ B) ∧ x ∈ (A ∪ 𝐶)) ↔ ((x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ 𝐶))) |
8 | 4, 7 | bitr2i 174 | . . 3 ⊢ (((x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ 𝐶)) ↔ x ∈ ((A ∪ B) ∩ (A ∪ 𝐶))) |
9 | 2, 3, 8 | 3bitri 195 | . 2 ⊢ ((x ∈ A ∨ x ∈ (B ∩ 𝐶)) ↔ x ∈ ((A ∪ B) ∩ (A ∪ 𝐶))) |
10 | 9 | uneqri 3079 | 1 ⊢ (A ∪ (B ∩ 𝐶)) = ((A ∪ B) ∩ (A ∪ 𝐶)) |
Colors of variables: wff set class |
Syntax hints: ∧ wa 97 ∨ wo 628 = wceq 1242 ∈ wcel 1390 ∪ cun 2909 ∩ cin 2910 |
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-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-un 2916 df-in 2918 |
This theorem is referenced by: undir 3181 |
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