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Theorem disjexc 33285
Description: A variant of disjex 33284, applicable for more generic families. (Contributed by Thierry Arnoux, 4-Oct-2016.)
Hypothesis
Ref Expression
disjexc.1 (𝑥 = 𝑦 → 𝐴 = 𝐵)
Assertion
Ref Expression
disjexc ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝑥 = 𝑦) → (𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅))
Distinct variable groups:   𝑧,𝐴   𝑧,𝐵
Allowed substitution hints:   𝐴(𝑥, 𝑦)   𝐵(𝑥, 𝑦)

Proof of Theorem disjexc
StepHypRef Expression
1 disjexc.1 . . 3 (𝑥 = 𝑦 → 𝐴 = 𝐵)
21imim2i 17 . 2 ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝑥 = 𝑦) → (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵))
3 orcom 884 . . 3 ((𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵))
4 df-in 4094 . . . . . . 7 (𝐴 ∩ 𝐵) = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)}
54neeq1i 3023 . . . . . 6 ((𝐴 ∩ 𝐵) ≠ ∅ ↔ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅)
6 abn0 4337 . . . . . 6 ({𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅ ↔ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))
75, 6bitr2i 279 . . . . 5 (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ↔ (𝐴 ∩ 𝐵) ≠ ∅)
87necon2bbii 3010 . . . 4 ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))
98orbi2i 926 . . 3 ((𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅) ↔ (𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)))
10 imor 867 . . 3 ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵))
113, 9, 103bitr4i 306 . 2 ((𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅) ↔ (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵))
122, 11sylibr 237 1 ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝑥 = 𝑦) → (𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∨ wo 861   = wceq 1570  ∃wex 1812   ∈ wcel 2147  {cab 2741   ≠ wne 2959   ∩ cin 4092  ∅c0 4281
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-ne 2960  df-dif 4050  df-in 4094  df-nul 4282
This theorem is used by: (None)
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