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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 3820 | . . . . . 6 ⊢ (𝐵 = ∅ → ∪ 𝐵 = ∪ ∅) | |
2 | uni0 3838 | . . . . . 6 ⊢ ∪ ∅ = ∅ | |
3 | 1, 2 | eqtrdi 2226 | . . . . 5 ⊢ (𝐵 = ∅ → ∪ 𝐵 = ∅) |
4 | 3 | uneq2d 3291 | . . . 4 ⊢ (𝐵 = ∅ → (𝐴 ∪ ∪ 𝐵) = (𝐴 ∪ ∅)) |
5 | un0 3458 | . . . 4 ⊢ (𝐴 ∪ ∅) = 𝐴 | |
6 | 4, 5 | eqtrdi 2226 | . . 3 ⊢ (𝐵 = ∅ → (𝐴 ∪ ∪ 𝐵) = 𝐴) |
7 | iuneq1 3901 | . . . 4 ⊢ (𝐵 = ∅ → ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = ∪ 𝑥 ∈ ∅ (𝐴 ∪ 𝑥)) | |
8 | 0iun 3946 | . . . 4 ⊢ ∪ 𝑥 ∈ ∅ (𝐴 ∪ 𝑥) = ∅ | |
9 | 7, 8 | eqtrdi 2226 | . . 3 ⊢ (𝐵 = ∅ → ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = ∅) |
10 | 6, 9 | eqeq12d 2192 | . 2 ⊢ (𝐵 = ∅ → ((𝐴 ∪ ∪ 𝐵) = ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) ↔ 𝐴 = ∅)) |
11 | 10 | biimpcd 159 | 1 ⊢ ((𝐴 ∪ ∪ 𝐵) = ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) → (𝐵 = ∅ → 𝐴 = ∅)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 = wceq 1353 ∪ cun 3129 ∅c0 3424 ∪ cuni 3811 ∪ ciun 3888 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-in1 614 ax-in2 615 ax-io 709 ax-5 1447 ax-7 1448 ax-gen 1449 ax-ie1 1493 ax-ie2 1494 ax-8 1504 ax-10 1505 ax-11 1506 ax-i12 1507 ax-bndl 1509 ax-4 1510 ax-17 1526 ax-i9 1530 ax-ial 1534 ax-i5r 1535 ax-ext 2159 |
This theorem depends on definitions: df-bi 117 df-tru 1356 df-fal 1359 df-nf 1461 df-sb 1763 df-clab 2164 df-cleq 2170 df-clel 2173 df-nfc 2308 df-ral 2460 df-rex 2461 df-v 2741 df-dif 3133 df-un 3135 df-in 3137 df-ss 3144 df-nul 3425 df-sn 3600 df-uni 3812 df-iun 3890 |
This theorem is referenced by: (None) |
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