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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 3684 | . . . . . 6 ⊢ (𝐵 = ∅ → ∪ 𝐵 = ∪ ∅) | |
2 | uni0 3702 | . . . . . 6 ⊢ ∪ ∅ = ∅ | |
3 | 1, 2 | syl6eq 2143 | . . . . 5 ⊢ (𝐵 = ∅ → ∪ 𝐵 = ∅) |
4 | 3 | uneq2d 3169 | . . . 4 ⊢ (𝐵 = ∅ → (𝐴 ∪ ∪ 𝐵) = (𝐴 ∪ ∅)) |
5 | un0 3335 | . . . 4 ⊢ (𝐴 ∪ ∅) = 𝐴 | |
6 | 4, 5 | syl6eq 2143 | . . 3 ⊢ (𝐵 = ∅ → (𝐴 ∪ ∪ 𝐵) = 𝐴) |
7 | iuneq1 3765 | . . . 4 ⊢ (𝐵 = ∅ → ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = ∪ 𝑥 ∈ ∅ (𝐴 ∪ 𝑥)) | |
8 | 0iun 3809 | . . . 4 ⊢ ∪ 𝑥 ∈ ∅ (𝐴 ∪ 𝑥) = ∅ | |
9 | 7, 8 | syl6eq 2143 | . . 3 ⊢ (𝐵 = ∅ → ∪ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = ∅) |
10 | 6, 9 | eqeq12d 2109 | . 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 1296 ∪ cun 3011 ∅c0 3302 ∪ cuni 3675 ∪ ciun 3752 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 582 ax-in2 583 ax-io 668 ax-5 1388 ax-7 1389 ax-gen 1390 ax-ie1 1434 ax-ie2 1435 ax-8 1447 ax-10 1448 ax-11 1449 ax-i12 1450 ax-bndl 1451 ax-4 1452 ax-17 1471 ax-i9 1475 ax-ial 1479 ax-i5r 1480 ax-ext 2077 |
This theorem depends on definitions: df-bi 116 df-tru 1299 df-fal 1302 df-nf 1402 df-sb 1700 df-clab 2082 df-cleq 2088 df-clel 2091 df-nfc 2224 df-ral 2375 df-rex 2376 df-v 2635 df-dif 3015 df-un 3017 df-in 3019 df-ss 3026 df-nul 3303 df-sn 3472 df-uni 3676 df-iun 3754 |
This theorem is referenced by: (None) |
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