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Theorem itg2lr 25722
Description: Sufficient condition for elementhood in the set 𝐿. (Contributed by Mario Carneiro, 28-Jun-2014.)
Hypothesis
Ref Expression
itg2val.1 𝐿 = {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔))}
Assertion
Ref Expression
itg2lr ((𝐺 ∈ dom ∫1𝐺r𝐹) → (∫1𝐺) ∈ 𝐿)
Distinct variable groups:   𝑥,𝑔,𝐹   𝑔,𝐺,𝑥
Allowed substitution hints:   𝐿(𝑥,𝑔)

Proof of Theorem itg2lr
StepHypRef Expression
1 eqid 2740 . . 3 (∫1𝐺) = (∫1𝐺)
2 breq1 5082 . . . . 5 (𝑔 = 𝐺 → (𝑔r𝐹𝐺r𝐹))
3 fveq2 6834 . . . . . 6 (𝑔 = 𝐺 → (∫1𝑔) = (∫1𝐺))
43eqeq2d 2751 . . . . 5 (𝑔 = 𝐺 → ((∫1𝐺) = (∫1𝑔) ↔ (∫1𝐺) = (∫1𝐺)))
52, 4anbi12d 638 . . . 4 (𝑔 = 𝐺 → ((𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)) ↔ (𝐺r𝐹 ∧ (∫1𝐺) = (∫1𝐺))))
65rspcev 3567 . . 3 ((𝐺 ∈ dom ∫1 ∧ (𝐺r𝐹 ∧ (∫1𝐺) = (∫1𝐺))) → ∃𝑔 ∈ dom ∫1(𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
71, 6mpanr2 710 . 2 ((𝐺 ∈ dom ∫1𝐺r𝐹) → ∃𝑔 ∈ dom ∫1(𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
8 itg2val.1 . . 3 𝐿 = {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔))}
98itg2l 25721 . 2 ((∫1𝐺) ∈ 𝐿 ↔ ∃𝑔 ∈ dom ∫1(𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
107, 9sylibr 235 1 ((𝐺 ∈ dom ∫1𝐺r𝐹) → (∫1𝐺) ∈ 𝐿)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 396   = wceq 1547  wcel 2119  {cab 2718  wrex 3064   class class class wbr 5079  dom cdm 5625  cfv 6492  r cofr 7626  cle 11178  1citg1 25607
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712  ax-nul 5235
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-ne 2936  df-ral 3055  df-rex 3065  df-rab 3393  df-v 3434  df-dif 3893  df-un 3895  df-ss 3907  df-nul 4269  df-if 4462  df-sn 4563  df-pr 4565  df-op 4569  df-uni 4846  df-br 5080  df-iota 6448  df-fv 6500
This theorem is referenced by:  itg2ub  25725
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