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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 871 | . 2 ⊢ ((𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵)) | |
| 2 | df-in 3958 | . . . . . 6 ⊢ (𝐴 ∩ 𝐵) = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} | |
| 3 | 2 | neeq1i 3005 | . . . . 5 ⊢ ((𝐴 ∩ 𝐵) ≠ ∅ ↔ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅) | 
| 4 | abn0 4385 | . . . . 5 ⊢ ({𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅ ↔ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) | |
| 5 | 3, 4 | bitr2i 276 | . . . 4 ⊢ (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ↔ (𝐴 ∩ 𝐵) ≠ ∅) | 
| 6 | 5 | necon2bbii 2992 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) | 
| 7 | 6 | orbi2i 913 | . 2 ⊢ ((𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅) ↔ (𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))) | 
| 8 | imor 854 | . 2 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵)) | |
| 9 | 1, 7, 8 | 3bitr4ri 304 | 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 206 ∧ wa 395 ∨ wo 848 = wceq 1540 ∃wex 1779 ∈ wcel 2108 {cab 2714 ≠ wne 2940 ∩ cin 3950 ∅c0 4333 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-9 2118 ax-10 2141 ax-11 2157 ax-12 2177 ax-ext 2708 | 
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 849 df-tru 1543 df-fal 1553 df-ex 1780 df-nf 1784 df-sb 2065 df-clab 2715 df-cleq 2729 df-ne 2941 df-dif 3954 df-in 3958 df-nul 4334 | 
| This theorem is referenced by: (None) | 
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