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Theorem disjnf 28766
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 Masza.) (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 3784 . . . 4 (𝐵𝐵) = 𝐵
21eqeq1i 2615 . . 3 ((𝐵𝐵) = ∅ ↔ 𝐵 = ∅)
32orbi1i 541 . 2 (((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦) ↔ (𝐵 = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
4 eqidd 2611 . . . 4 (𝑥 = 𝑦𝐵 = 𝐵)
54disjor 4567 . . 3 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅))
6 orcom 401 . . . . . 6 ((𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦))
76ralbii 2963 . . . . 5 (∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ∀𝑦𝐴 ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦))
8 r19.32v 3064 . . . . 5 (∀𝑦𝐴 ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
97, 8bitri 263 . . . 4 (∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
109ralbii 2963 . . 3 (∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ∀𝑥𝐴 ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
11 r19.32v 3064 . . 3 (∀𝑥𝐴 ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
125, 10, 113bitri 285 . 2 (Disj 𝑥𝐴 𝐵 ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
13 moel 28707 . . 3 (∃*𝑥 𝑥𝐴 ↔ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦)
1413orbi2i 540 . 2 ((𝐵 = ∅ ∨ ∃*𝑥 𝑥𝐴) ↔ (𝐵 = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
153, 12, 143bitr4i 291 1 (Disj 𝑥𝐴 𝐵 ↔ (𝐵 = ∅ ∨ ∃*𝑥 𝑥𝐴))
Colors of variables: wff setvar class
Syntax hints:  wb 195  wo 382   = wceq 1475  wcel 1977  ∃*wmo 2459  wral 2896  cin 3539  c0 3874  Disj wdisj 4553
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590
This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-eu 2462  df-mo 2463  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ral 2901  df-rmo 2904  df-v 3175  df-dif 3543  df-in 3547  df-nul 3875  df-disj 4554
This theorem is referenced by: (None)
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