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Theorem disjnf 32915
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 4179 . . . 4 (𝐵𝐵) = 𝐵
21eqeq1i 2768 . . 3 ((𝐵𝐵) = ∅ ↔ 𝐵 = ∅)
32orbi1i 926 . 2 (((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦) ↔ (𝐵 = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
4 eqidd 2764 . . . 4 (𝑥 = 𝑦𝐵 = 𝐵)
54disjor 5091 . . 3 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅))
6 orcom 883 . . . . . 6 ((𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦))
76ralbii 3111 . . . . 5 (∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ∀𝑦𝐴 ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦))
8 r19.32v 3198 . . . . 5 (∀𝑦𝐴 ((𝐵𝐵) = ∅ ∨ 𝑥 = 𝑦) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
97, 8bitri 278 . . . 4 (∀𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
109ralbii 3111 . . 3 (∀𝑥𝐴𝑦𝐴 (𝑥 = 𝑦 ∨ (𝐵𝐵) = ∅) ↔ ∀𝑥𝐴 ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦))
11 r19.32v 3198 . . 3 (∀𝑥𝐴 ((𝐵𝐵) = ∅ ∨ ∀𝑦𝐴 𝑥 = 𝑦) ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
125, 10, 113bitri 300 . 2 (Disj 𝑥𝐴 𝐵 ↔ ((𝐵𝐵) = ∅ ∨ ∀𝑥𝐴𝑦𝐴 𝑥 = 𝑦))
13 moel 3389 . . 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 1570  wcel 2143  ∃*wmo 2565  wral 3079  cin 3904  c0 4286  Disj wdisj 5076
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-11 2192  ax-ext 2735
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-mo 2567  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-rmo 3369  df-v 3457  df-dif 3908  df-in 3912  df-nul 4287  df-disj 5077
This theorem is referenced by: (None)
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