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Theorem itg2l 25698
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 2829 . 2 (𝐴𝐿𝐴 ∈ {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔))})
3 simpr 484 . . . . 5 ((𝑔r𝐹𝐴 = (∫1𝑔)) → 𝐴 = (∫1𝑔))
4 fvex 6855 . . . . 5 (∫1𝑔) ∈ V
53, 4eqeltrdi 2845 . . . 4 ((𝑔r𝐹𝐴 = (∫1𝑔)) → 𝐴 ∈ V)
65rexlimivw 3135 . . 3 (∃𝑔 ∈ dom ∫1(𝑔r𝐹𝐴 = (∫1𝑔)) → 𝐴 ∈ V)
7 eqeq1 2741 . . . . 5 (𝑥 = 𝐴 → (𝑥 = (∫1𝑔) ↔ 𝐴 = (∫1𝑔)))
87anbi2d 631 . . . 4 (𝑥 = 𝐴 → ((𝑔r𝐹𝑥 = (∫1𝑔)) ↔ (𝑔r𝐹𝐴 = (∫1𝑔))))
98rexbidv 3162 . . 3 (𝑥 = 𝐴 → (∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔)) ↔ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝐴 = (∫1𝑔))))
106, 9elab3 3643 . 2 (𝐴 ∈ {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔))} ↔ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝐴 = (∫1𝑔)))
112, 10bitri 275 1 (𝐴𝐿 ↔ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝐴 = (∫1𝑔)))
Colors of variables: wff setvar class
Syntax hints:  wb 206  wa 395   = wceq 1542  wcel 2114  {cab 2715  wrex 3062  Vcvv 3442   class class class wbr 5100  dom cdm 5632  cfv 6500  r cofr 7631  cle 11179  1citg1 25584
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 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-nul 5253
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ne 2934  df-rex 3063  df-v 3444  df-dif 3906  df-un 3908  df-ss 3920  df-nul 4288  df-sn 4583  df-pr 4585  df-uni 4866  df-iota 6456  df-fv 6508
This theorem is referenced by:  itg2lr  25699
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