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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 3803 | . . . . . 6 ⊢ (𝐵 = ∅ → ∪ 𝐵 = ∪ ∅) | |
2 | uni0 3821 | . . . . . 6 ⊢ ∪ ∅ = ∅ | |
3 | 1, 2 | eqtrdi 2219 | . . . . 5 ⊢ (𝐵 = ∅ → ∪ 𝐵 = ∅) |
4 | 3 | uneq2d 3281 | . . . 4 ⊢ (𝐵 = ∅ → (𝐴 ∪ ∪ 𝐵) = (𝐴 ∪ ∅)) |
5 | un0 3447 | . . . 4 ⊢ (𝐴 ∪ ∅) = 𝐴 | |
6 | 4, 5 | eqtrdi 2219 | . . 3 ⊢ (𝐵 = ∅ → (𝐴 ∪ ∪ 𝐵) = 𝐴) |
7 | iuneq1 3884 | . . . 4 ⊢ (𝐵 = ∅ → ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = ∪ 𝑥 ∈ ∅ (𝐴 ∪ 𝑥)) | |
8 | 0iun 3928 | . . . 4 ⊢ ∪ 𝑥 ∈ ∅ (𝐴 ∪ 𝑥) = ∅ | |
9 | 7, 8 | eqtrdi 2219 | . . 3 ⊢ (𝐵 = ∅ → ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = ∅) |
10 | 6, 9 | eqeq12d 2185 | . 2 ⊢ (𝐵 = ∅ → ((𝐴 ∪ ∪ 𝐵) = ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) ↔ 𝐴 = ∅)) |
11 | 10 | biimpcd 158 | 1 ⊢ ((𝐴 ∪ ∪ 𝐵) = ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) → (𝐵 = ∅ → 𝐴 = ∅)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 = wceq 1348 ∪ cun 3119 ∅c0 3414 ∪ cuni 3794 ∪ ciun 3871 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 609 ax-in2 610 ax-io 704 ax-5 1440 ax-7 1441 ax-gen 1442 ax-ie1 1486 ax-ie2 1487 ax-8 1497 ax-10 1498 ax-11 1499 ax-i12 1500 ax-bndl 1502 ax-4 1503 ax-17 1519 ax-i9 1523 ax-ial 1527 ax-i5r 1528 ax-ext 2152 |
This theorem depends on definitions: df-bi 116 df-tru 1351 df-fal 1354 df-nf 1454 df-sb 1756 df-clab 2157 df-cleq 2163 df-clel 2166 df-nfc 2301 df-ral 2453 df-rex 2454 df-v 2732 df-dif 3123 df-un 3125 df-in 3127 df-ss 3134 df-nul 3415 df-sn 3587 df-uni 3795 df-iun 3873 |
This theorem is referenced by: (None) |
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