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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 3833 | . . . . . 6 ⊢ (𝐵 = ∅ → ∪ 𝐵 = ∪ ∅) | |
2 | uni0 3851 | . . . . . 6 ⊢ ∪ ∅ = ∅ | |
3 | 1, 2 | eqtrdi 2238 | . . . . 5 ⊢ (𝐵 = ∅ → ∪ 𝐵 = ∅) |
4 | 3 | uneq2d 3304 | . . . 4 ⊢ (𝐵 = ∅ → (𝐴 ∪ ∪ 𝐵) = (𝐴 ∪ ∅)) |
5 | un0 3471 | . . . 4 ⊢ (𝐴 ∪ ∅) = 𝐴 | |
6 | 4, 5 | eqtrdi 2238 | . . 3 ⊢ (𝐵 = ∅ → (𝐴 ∪ ∪ 𝐵) = 𝐴) |
7 | iuneq1 3914 | . . . 4 ⊢ (𝐵 = ∅ → ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = ∪ 𝑥 ∈ ∅ (𝐴 ∪ 𝑥)) | |
8 | 0iun 3959 | . . . 4 ⊢ ∪ 𝑥 ∈ ∅ (𝐴 ∪ 𝑥) = ∅ | |
9 | 7, 8 | eqtrdi 2238 | . . 3 ⊢ (𝐵 = ∅ → ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = ∅) |
10 | 6, 9 | eqeq12d 2204 | . 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 1364 ∪ cun 3142 ∅c0 3437 ∪ cuni 3824 ∪ ciun 3901 |
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 615 ax-in2 616 ax-io 710 ax-5 1458 ax-7 1459 ax-gen 1460 ax-ie1 1504 ax-ie2 1505 ax-8 1515 ax-10 1516 ax-11 1517 ax-i12 1518 ax-bndl 1520 ax-4 1521 ax-17 1537 ax-i9 1541 ax-ial 1545 ax-i5r 1546 ax-ext 2171 |
This theorem depends on definitions: df-bi 117 df-tru 1367 df-fal 1370 df-nf 1472 df-sb 1774 df-clab 2176 df-cleq 2182 df-clel 2185 df-nfc 2321 df-ral 2473 df-rex 2474 df-v 2754 df-dif 3146 df-un 3148 df-in 3150 df-ss 3157 df-nul 3438 df-sn 3613 df-uni 3825 df-iun 3903 |
This theorem is referenced by: (None) |
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