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 866 | . 2 ⊢ ((𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵)) | |
2 | df-in 3942 | . . . . . 6 ⊢ (𝐴 ∩ 𝐵) = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} | |
3 | 2 | neeq1i 3080 | . . . . 5 ⊢ ((𝐴 ∩ 𝐵) ≠ ∅ ↔ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅) |
4 | abn0 4335 | . . . . 5 ⊢ ({𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅ ↔ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) | |
5 | 3, 4 | bitr2i 278 | . . . 4 ⊢ (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ↔ (𝐴 ∩ 𝐵) ≠ ∅) |
6 | 5 | necon2bbii 3067 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) |
7 | 6 | orbi2i 909 | . 2 ⊢ ((𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅) ↔ (𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))) |
8 | imor 849 | . 2 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵)) | |
9 | 1, 7, 8 | 3bitr4ri 306 | 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 208 ∧ wa 398 ∨ wo 843 = wceq 1533 ∃wex 1776 ∈ wcel 2110 {cab 2799 ≠ wne 3016 ∩ cin 3934 ∅c0 4290 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1907 ax-6 1966 ax-7 2011 ax-8 2112 ax-9 2120 ax-10 2141 ax-11 2157 ax-12 2173 ax-ext 2793 |
This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-tru 1536 df-ex 1777 df-nf 1781 df-sb 2066 df-clab 2800 df-cleq 2814 df-clel 2893 df-nfc 2963 df-ne 3017 df-dif 3938 df-in 3942 df-nul 4291 |
This theorem is referenced by: (None) |
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