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Theorem disjex 32612
Description: Two ways to say that two classes are disjoint (or equal). (Contributed by Thierry Arnoux, 4-Oct-2016.)
Assertion
Ref Expression
disjex ((∃𝑧(𝑧𝐴𝑧𝐵) → 𝐴 = 𝐵) ↔ (𝐴 = 𝐵 ∨ (𝐴𝐵) = ∅))
Distinct variable groups:   𝑧,𝐴   𝑧,𝐵

Proof of Theorem disjex
StepHypRef Expression
1 orcom 870 . 2 ((𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧𝐴𝑧𝐵)) ↔ (¬ ∃𝑧(𝑧𝐴𝑧𝐵) ∨ 𝐴 = 𝐵))
2 df-in 3970 . . . . . 6 (𝐴𝐵) = {𝑧 ∣ (𝑧𝐴𝑧𝐵)}
32neeq1i 3003 . . . . 5 ((𝐴𝐵) ≠ ∅ ↔ {𝑧 ∣ (𝑧𝐴𝑧𝐵)} ≠ ∅)
4 abn0 4391 . . . . 5 ({𝑧 ∣ (𝑧𝐴𝑧𝐵)} ≠ ∅ ↔ ∃𝑧(𝑧𝐴𝑧𝐵))
53, 4bitr2i 276 . . . 4 (∃𝑧(𝑧𝐴𝑧𝐵) ↔ (𝐴𝐵) ≠ ∅)
65necon2bbii 2990 . . 3 ((𝐴𝐵) = ∅ ↔ ¬ ∃𝑧(𝑧𝐴𝑧𝐵))
76orbi2i 912 . 2 ((𝐴 = 𝐵 ∨ (𝐴𝐵) = ∅) ↔ (𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧𝐴𝑧𝐵)))
8 imor 853 . 2 ((∃𝑧(𝑧𝐴𝑧𝐵) → 𝐴 = 𝐵) ↔ (¬ ∃𝑧(𝑧𝐴𝑧𝐵) ∨ 𝐴 = 𝐵))
91, 7, 83bitr4ri 304 1 ((∃𝑧(𝑧𝐴𝑧𝐵) → 𝐴 = 𝐵) ↔ (𝐴 = 𝐵 ∨ (𝐴𝐵) = ∅))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 206  wa 395  wo 847   = wceq 1537  wex 1776  wcel 2106  {cab 2712  wne 2938  cin 3962  c0 4339
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 1908  ax-6 1965  ax-7 2005  ax-9 2116  ax-10 2139  ax-11 2155  ax-12 2175  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-fal 1550  df-ex 1777  df-nf 1781  df-sb 2063  df-clab 2713  df-cleq 2727  df-ne 2939  df-dif 3966  df-in 3970  df-nul 4340
This theorem is referenced by: (None)
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