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Theorem ndisj2 45812
Description: A non-disjointness condition. (Contributed by Glauco Siliprandi, 17-Aug-2020.)
Hypothesis
Ref Expression
ndisj2.1 (𝑥 = 𝑦𝐵 = 𝐶)
Assertion
Ref Expression
ndisj2 Disj 𝑥𝐴 𝐵 ↔ ∃𝑥𝐴𝑦𝐴 (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
Distinct variable groups:   𝑥,𝐴,𝑦   𝑦,𝐵   𝑥,𝐶
Allowed substitution hints:   𝐵(𝑥)   𝐶(𝑦)

Proof of Theorem ndisj2
StepHypRef Expression
1 ndisj2.1 . . . 4 (𝑥 = 𝑦𝐵 = 𝐶)
21disjor 5096 . . 3 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅))
32notbii 323 . 2 Disj 𝑥𝐴 𝐵 ↔ ¬ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅))
4 rexnal 3120 . 2 (∃𝑥𝐴 ¬ ∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ ¬ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅))
5 rexnal 3120 . . . 4 (∃𝑦𝐴 ¬ (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ ¬ ∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅))
6 ioran 999 . . . . . 6 (¬ (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ (¬ 𝑥 = 𝑦 ∧ ¬ (𝐵𝐶) = ∅))
7 df-ne 2962 . . . . . . 7 (𝑥𝑦 ↔ ¬ 𝑥 = 𝑦)
8 df-ne 2962 . . . . . . 7 ((𝐵𝐶) ≠ ∅ ↔ ¬ (𝐵𝐶) = ∅)
97, 8anbi12i 640 . . . . . 6 ((𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅) ↔ (¬ 𝑥 = 𝑦 ∧ ¬ (𝐵𝐶) = ∅))
106, 9bitr4i 281 . . . . 5 (¬ (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
1110rexbii 3115 . . . 4 (∃𝑦𝐴 ¬ (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ ∃𝑦𝐴 (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
125, 11bitr3i 280 . . 3 (¬ ∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ ∃𝑦𝐴 (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
1312rexbii 3115 . 2 (∃𝑥𝐴 ¬ ∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ ∃𝑥𝐴𝑦𝐴 (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
143, 4, 133bitr2i 302 1 Disj 𝑥𝐴 𝐵 ↔ ∃𝑥𝐴𝑦𝐴 (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3  wb 209  wa 401  wo 861   = wceq 1570  wne 2961  wral 3082  wrex 3092  cin 3907  c0 4289  Disj wdisj 5081
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 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-11 2195  ax-ext 2738
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-mo 2570  df-clab 2745  df-cleq 2758  df-clel 2841  df-ne 2962  df-ral 3083  df-rex 3093  df-rmo 3372  df-v 3460  df-dif 3911  df-in 3915  df-nul 4290  df-disj 5082
This theorem is used by:  disjrnmpt2  45947
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