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Theorem ndisj2 40772
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 4916 . . 3 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅))
32notbii 312 . 2 Disj 𝑥𝐴 𝐵 ↔ ¬ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅))
4 rexnal 3187 . 2 (∃𝑥𝐴 ¬ ∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ ¬ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅))
5 rexnal 3187 . . . 4 (∃𝑦𝐴 ¬ (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ ¬ ∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅))
6 ioran 967 . . . . . 6 (¬ (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ (¬ 𝑥 = 𝑦 ∧ ¬ (𝐵𝐶) = ∅))
7 df-ne 2970 . . . . . . 7 (𝑥𝑦 ↔ ¬ 𝑥 = 𝑦)
8 df-ne 2970 . . . . . . 7 ((𝐵𝐶) ≠ ∅ ↔ ¬ (𝐵𝐶) = ∅)
97, 8anbi12i 618 . . . . . 6 ((𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅) ↔ (¬ 𝑥 = 𝑦 ∧ ¬ (𝐵𝐶) = ∅))
106, 9bitr4i 270 . . . . 5 (¬ (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
1110rexbii 3196 . . . 4 (∃𝑦𝐴 ¬ (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ ∃𝑦𝐴 (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
125, 11bitr3i 269 . . 3 (¬ ∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ ∃𝑦𝐴 (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
1312rexbii 3196 . 2 (∃𝑥𝐴 ¬ ∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐶) = ∅) ↔ ∃𝑥𝐴𝑦𝐴 (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
143, 4, 133bitr2i 291 1 Disj 𝑥𝐴 𝐵 ↔ ∃𝑥𝐴𝑦𝐴 (𝑥𝑦 ∧ (𝐵𝐶) ≠ ∅))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 198  wa 387  wo 834   = wceq 1508  wne 2969  wral 3090  wrex 3091  cin 3830  c0 4181  Disj wdisj 4902
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1759  ax-4 1773  ax-5 1870  ax-6 1929  ax-7 1966  ax-8 2053  ax-9 2060  ax-10 2080  ax-11 2094  ax-12 2107  ax-ext 2752
This theorem depends on definitions:  df-bi 199  df-an 388  df-or 835  df-tru 1511  df-ex 1744  df-nf 1748  df-sb 2017  df-mo 2551  df-clab 2761  df-cleq 2773  df-clel 2848  df-nfc 2920  df-ne 2970  df-ral 3095  df-rex 3096  df-rmo 3098  df-v 3419  df-dif 3834  df-in 3838  df-nul 4182  df-disj 4903
This theorem is referenced by:  disjrnmpt2  40910
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