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Mirrors > Home > ILE Home > Th. List > iununir | GIF version |
Description: A relationship involving union and indexed union. Exercise 25 of [Enderton] p. 33 but with biconditional changed to implication. (Contributed by Jim Kingdon, 19-Aug-2018.) |
Ref | Expression |
---|---|
iununir | ⊢ ((𝐴 ∪ ∪ 𝐵) = ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) → (𝐵 = ∅ → 𝐴 = ∅)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | unieq 3610 | . . . . . 6 ⊢ (𝐵 = ∅ → ∪ 𝐵 = ∪ ∅) | |
2 | uni0 3628 | . . . . . 6 ⊢ ∪ ∅ = ∅ | |
3 | 1, 2 | syl6eq 2129 | . . . . 5 ⊢ (𝐵 = ∅ → ∪ 𝐵 = ∅) |
4 | 3 | uneq2d 3126 | . . . 4 ⊢ (𝐵 = ∅ → (𝐴 ∪ ∪ 𝐵) = (𝐴 ∪ ∅)) |
5 | un0 3278 | . . . 4 ⊢ (𝐴 ∪ ∅) = 𝐴 | |
6 | 4, 5 | syl6eq 2129 | . . 3 ⊢ (𝐵 = ∅ → (𝐴 ∪ ∪ 𝐵) = 𝐴) |
7 | iuneq1 3691 | . . . 4 ⊢ (𝐵 = ∅ → ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = ∪ 𝑥 ∈ ∅ (𝐴 ∪ 𝑥)) | |
8 | 0iun 3735 | . . . 4 ⊢ ∪ 𝑥 ∈ ∅ (𝐴 ∪ 𝑥) = ∅ | |
9 | 7, 8 | syl6eq 2129 | . . 3 ⊢ (𝐵 = ∅ → ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = ∅) |
10 | 6, 9 | eqeq12d 2095 | . 2 ⊢ (𝐵 = ∅ → ((𝐴 ∪ ∪ 𝐵) = ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) ↔ 𝐴 = ∅)) |
11 | 10 | biimpcd 157 | 1 ⊢ ((𝐴 ∪ ∪ 𝐵) = ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) → (𝐵 = ∅ → 𝐴 = ∅)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 = wceq 1284 ∪ cun 2971 ∅c0 3251 ∪ cuni 3601 ∪ ciun 3678 |
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-in1 576 ax-in2 577 ax-io 662 ax-5 1376 ax-7 1377 ax-gen 1378 ax-ie1 1422 ax-ie2 1423 ax-8 1435 ax-10 1436 ax-11 1437 ax-i12 1438 ax-bndl 1439 ax-4 1440 ax-17 1459 ax-i9 1463 ax-ial 1467 ax-i5r 1468 ax-ext 2063 |
This theorem depends on definitions: df-bi 115 df-tru 1287 df-fal 1290 df-nf 1390 df-sb 1686 df-clab 2068 df-cleq 2074 df-clel 2077 df-nfc 2208 df-ral 2353 df-rex 2354 df-v 2603 df-dif 2975 df-un 2977 df-in 2979 df-ss 2986 df-nul 3252 df-sn 3404 df-uni 3602 df-iun 3680 |
This theorem is referenced by: (None) |
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