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Theorem itg2lr 24883
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 2738 . . 3 (∫1𝐺) = (∫1𝐺)
2 breq1 5077 . . . . 5 (𝑔 = 𝐺 → (𝑔r𝐹𝐺r𝐹))
3 fveq2 6767 . . . . . 6 (𝑔 = 𝐺 → (∫1𝑔) = (∫1𝐺))
43eqeq2d 2749 . . . . 5 (𝑔 = 𝐺 → ((∫1𝐺) = (∫1𝑔) ↔ (∫1𝐺) = (∫1𝐺)))
52, 4anbi12d 631 . . . 4 (𝑔 = 𝐺 → ((𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)) ↔ (𝐺r𝐹 ∧ (∫1𝐺) = (∫1𝐺))))
65rspcev 3560 . . 3 ((𝐺 ∈ dom ∫1 ∧ (𝐺r𝐹 ∧ (∫1𝐺) = (∫1𝐺))) → ∃𝑔 ∈ dom ∫1(𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
71, 6mpanr2 701 . 2 ((𝐺 ∈ dom ∫1𝐺r𝐹) → ∃𝑔 ∈ dom ∫1(𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
8 itg2val.1 . . 3 𝐿 = {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔))}
98itg2l 24882 . 2 ((∫1𝐺) ∈ 𝐿 ↔ ∃𝑔 ∈ dom ∫1(𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
107, 9sylibr 233 1 ((𝐺 ∈ dom ∫1𝐺r𝐹) → (∫1𝐺) ∈ 𝐿)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 396   = wceq 1539  wcel 2106  {cab 2715  wrex 3065   class class class wbr 5074  dom cdm 5585  cfv 6427  r cofr 7523  cle 10998  1citg1 24767
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2709  ax-nul 5229
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2068  df-mo 2540  df-eu 2569  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rex 3070  df-rab 3073  df-v 3432  df-dif 3890  df-un 3892  df-in 3894  df-ss 3904  df-nul 4258  df-if 4461  df-sn 4563  df-pr 4565  df-op 4569  df-uni 4841  df-br 5075  df-iota 6385  df-fv 6435
This theorem is referenced by:  itg2ub  24886
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