Mathbox for Thierry Arnoux |
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Mirrors > Home > MPE Home > Th. List > Mathboxes > disjex | Structured version Visualization version GIF version |
Description: Two ways to say that two classes are disjoint (or equal). (Contributed by Thierry Arnoux, 4-Oct-2016.) |
Ref | Expression |
---|---|
disjex | ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | orcom 401 | . 2 ⊢ ((𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵)) | |
2 | df-in 3547 | . . . . . 6 ⊢ (𝐴 ∩ 𝐵) = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} | |
3 | 2 | neeq1i 2846 | . . . . 5 ⊢ ((𝐴 ∩ 𝐵) ≠ ∅ ↔ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅) |
4 | abn0 3908 | . . . . 5 ⊢ ({𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅ ↔ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) | |
5 | 3, 4 | bitr2i 264 | . . . 4 ⊢ (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ↔ (𝐴 ∩ 𝐵) ≠ ∅) |
6 | 5 | necon2bbii 2833 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) |
7 | 6 | orbi2i 540 | . 2 ⊢ ((𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅) ↔ (𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))) |
8 | imor 427 | . 2 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵)) | |
9 | 1, 7, 8 | 3bitr4ri 292 | 1 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅)) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ↔ wb 195 ∨ wo 382 ∧ wa 383 = wceq 1475 ∃wex 1695 ∈ wcel 1977 {cab 2596 ≠ wne 2780 ∩ cin 3539 ∅c0 3874 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-v 3175 df-dif 3543 df-in 3547 df-nul 3875 |
This theorem is referenced by: (None) |
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