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Theorem itg2l 25925
Description: Elementhood in the set 𝐿 of lower sums of the integral. (Contributed by Mario Carneiro, 28-Jun-2014.)
Hypothesis
Ref Expression
itg2val.1 𝐿 = {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔))}
Assertion
Ref Expression
itg2l (𝐴𝐿 ↔ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝐴 = (∫1𝑔)))
Distinct variable groups:   𝑥,𝑔,𝐴   𝑔,𝐹,𝑥
Allowed substitution hints:   𝐿(𝑥, 𝑔)

Proof of Theorem itg2l
StepHypRef Expression
1 itg2val.1 . . 3 𝐿 = {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔))}
21eleq2i 2858 . 2 (𝐴𝐿𝐴 ∈ {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔))})
3 simpr 490 . . . . 5 ((𝑔r𝐹𝐴 = (∫1𝑔)) → 𝐴 = (∫1𝑔))
4 fvex 6901 . . . . 5 (∫1𝑔) ∈ V
53, 4eqeltrdi 2874 . . . 4 ((𝑔r𝐹𝐴 = (∫1𝑔)) → 𝐴 ∈ V)
65rexlimivw 3165 . . 3 (∃𝑔 ∈ dom ∫1(𝑔r𝐹𝐴 = (∫1𝑔)) → 𝐴 ∈ V)
7 eqeq1 2770 . . . . 5 (𝑥 = 𝐴 → (𝑥 = (∫1𝑔) ↔ 𝐴 = (∫1𝑔)))
87anbi2d 642 . . . 4 (𝑥 = 𝐴 → ((𝑔r𝐹𝑥 = (∫1𝑔)) ↔ (𝑔r𝐹𝐴 = (∫1𝑔))))
98rexbidv 3192 . . 3 (𝑥 = 𝐴 → (∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔)) ↔ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝐴 = (∫1𝑔))))
106, 9elab3 3648 . 2 (𝐴 ∈ {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔))} ↔ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝐴 = (∫1𝑔)))
112, 10bitri 278 1 (𝐴𝐿 ↔ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝐴 = (∫1𝑔)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wb 209  wa 401   = wceq 1570  wcel 2146  {cab 2744  wrex 3092  Vcvv 3458   class class class wbr 5114  dom cdm 5666  cfv 6543  r cofr 7686  cle 11262  1citg1 25811
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738  ax-nul 5274
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ne 2962  df-rex 3093  df-v 3460  df-dif 3911  df-un 3913  df-ss 3925  df-nul 4290  df-sn 4595  df-pr 4597  df-uni 4878  df-iota 6499  df-fv 6551
This theorem is used by:  itg2lr  25926
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