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Theorem disjnf 32856
Description: In case 𝑥 is not free in 𝐵, disjointness is not so interesting since it reduces to cases where 𝐴 is a singleton. (Google Groups discussion with Peter Mazsa.) (Contributed by Thierry Arnoux, 26-Jul-2018.)
Assertion
Ref Expression
disjnf (Disj 𝑥𝐴 𝐵 ↔ (𝐵 = ∅ ∨ ∃*𝑥 𝑥𝐴))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem disjnf
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 inidm 4187 . . . 4 (𝐵𝐵) = 𝐵
21eqeq1i 2774 . . 3 ((𝐵𝐵) = ∅ ↔ 𝐵 = ∅)
32orbi1i 926 . 2 (((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦) ↔ (𝐵 = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
4 eqidd 2770 . . . 4 (𝑥 = 𝑦𝐵 = 𝐵)
54disjor 5095 . . 3 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅))
6 orcom 883 . . . . . 6 ((𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦))
76ralbii 3117 . . . . 5 (∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ∀𝑦𝐴 ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦))
8 r19.32v 3204 . . . . 5 (∀𝑦𝐴 ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
97, 8bitri 278 . . . 4 (∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
109ralbii 3117 . . 3 (∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ∀𝑥𝐴 ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
11 r19.32v 3204 . . 3 (∀𝑥𝐴 ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
125, 10, 113bitri 300 . 2 (Disj 𝑥𝐴 𝐵 ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
13 moel 3396 . . 3 (∃*𝑥 𝑥𝐴 ↔ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦)
1413orbi2i 925 . 2 ((𝐵 = ∅ ∨ ∃*𝑥 𝑥𝐴) ↔ (𝐵 = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
153, 12, 143bitr4i 306 1 (Disj 𝑥𝐴 𝐵 ↔ (𝐵 = ∅ ∨ ∃*𝑥 𝑥𝐴))
Colors of variables: wff setvar class
Syntax hints:  wb 209  wo 860   = wceq 1567  wcel 2149  ∃*wmo 2571  wral 3085  cin 3912  c0 4294  Disj wdisj 5080
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-11 2198  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-mo 2573  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-rmo 3376  df-v 3465  df-dif 3916  df-in 3920  df-nul 4295  df-disj 5081
This theorem is referenced by: (None)
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