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Theorem disjnf 30333
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 4145 . . . 4 (𝐵𝐵) = 𝐵
21eqeq1i 2803 . . 3 ((𝐵𝐵) = ∅ ↔ 𝐵 = ∅)
32orbi1i 911 . 2 (((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦) ↔ (𝐵 = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
4 eqidd 2799 . . . 4 (𝑥 = 𝑦𝐵 = 𝐵)
54disjor 5010 . . 3 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅))
6 orcom 867 . . . . . 6 ((𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦))
76ralbii 3133 . . . . 5 (∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ∀𝑦𝐴 ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦))
8 r19.32v 3294 . . . . 5 (∀𝑦𝐴 ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
97, 8bitri 278 . . . 4 (∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
109ralbii 3133 . . 3 (∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ∀𝑥𝐴 ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
11 r19.32v 3294 . . 3 (∀𝑥𝐴 ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
125, 10, 113bitri 300 . 2 (Disj 𝑥𝐴 𝐵 ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
13 moel 30259 . . 3 (∃*𝑥 𝑥𝐴 ↔ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦)
1413orbi2i 910 . 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 844   = wceq 1538  wcel 2111  ∃*wmo 2596  wral 3106  cin 3880  c0 4243  Disj wdisj 4995
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-11 2158  ax-ext 2770
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2070  df-mo 2598  df-clab 2777  df-cleq 2791  df-clel 2870  df-ral 3111  df-rmo 3114  df-v 3443  df-dif 3884  df-in 3888  df-nul 4244  df-disj 4996
This theorem is referenced by: (None)
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