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Theorem disjex 32605
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 871 . 2 ((𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧𝐴𝑧𝐵)) ↔ (¬ ∃𝑧(𝑧𝐴𝑧𝐵) ∨ 𝐴 = 𝐵))
2 df-in 3958 . . . . . 6 (𝐴𝐵) = {𝑧 ∣ (𝑧𝐴𝑧𝐵)}
32neeq1i 3005 . . . . 5 ((𝐴𝐵) ≠ ∅ ↔ {𝑧 ∣ (𝑧𝐴𝑧𝐵)} ≠ ∅)
4 abn0 4385 . . . . 5 ({𝑧 ∣ (𝑧𝐴𝑧𝐵)} ≠ ∅ ↔ ∃𝑧(𝑧𝐴𝑧𝐵))
53, 4bitr2i 276 . . . 4 (∃𝑧(𝑧𝐴𝑧𝐵) ↔ (𝐴𝐵) ≠ ∅)
65necon2bbii 2992 . . 3 ((𝐴𝐵) = ∅ ↔ ¬ ∃𝑧(𝑧𝐴𝑧𝐵))
76orbi2i 913 . 2 ((𝐴 = 𝐵 ∨ (𝐴𝐵) = ∅) ↔ (𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧𝐴𝑧𝐵)))
8 imor 854 . 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 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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