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Theorem disjnf 29947
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 4043 . . . 4 (𝐵𝐵) = 𝐵
21eqeq1i 2783 . . 3 ((𝐵𝐵) = ∅ ↔ 𝐵 = ∅)
32orbi1i 900 . 2 (((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦) ↔ (𝐵 = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
4 eqidd 2779 . . . 4 (𝑥 = 𝑦𝐵 = 𝐵)
54disjor 4868 . . 3 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅))
6 orcom 859 . . . . . 6 ((𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦))
76ralbii 3162 . . . . 5 (∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ∀𝑦𝐴 ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦))
8 r19.32v 3269 . . . . 5 (∀𝑦𝐴 ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
97, 8bitri 267 . . . 4 (∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
109ralbii 3162 . . 3 (∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ∀𝑥𝐴 ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
11 r19.32v 3269 . . 3 (∀𝑥𝐴 ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
125, 10, 113bitri 289 . 2 (Disj 𝑥𝐴 𝐵 ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
13 moel 29895 . . 3 (∃*𝑥 𝑥𝐴 ↔ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦)
1413orbi2i 899 . 2 ((𝐵 = ∅ ∨ ∃*𝑥 𝑥𝐴) ↔ (𝐵 = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
153, 12, 143bitr4i 295 1 (Disj 𝑥𝐴 𝐵 ↔ (𝐵 = ∅ ∨ ∃*𝑥 𝑥𝐴))
Colors of variables: wff setvar class
Syntax hints:  wb 198  wo 836   = wceq 1601  wcel 2107  ∃*wmo 2549  wral 3090  cin 3791  c0 4141  Disj wdisj 4854
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2055  ax-9 2116  ax-10 2135  ax-11 2150  ax-12 2163  ax-13 2334  ax-ext 2754
This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-tru 1605  df-ex 1824  df-nf 1828  df-sb 2012  df-mo 2551  df-clab 2764  df-cleq 2770  df-clel 2774  df-nfc 2921  df-ral 3095  df-rmo 3098  df-v 3400  df-dif 3795  df-in 3799  df-nul 4142  df-disj 4855
This theorem is referenced by: (None)
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