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Theorem itg2lr 25680
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 2728 . . 3 (∫1𝐺) = (∫1𝐺)
2 breq1 5155 . . . . 5 (𝑔 = 𝐺 → (𝑔r𝐹𝐺r𝐹))
3 fveq2 6902 . . . . . 6 (𝑔 = 𝐺 → (∫1𝑔) = (∫1𝐺))
43eqeq2d 2739 . . . . 5 (𝑔 = 𝐺 → ((∫1𝐺) = (∫1𝑔) ↔ (∫1𝐺) = (∫1𝐺)))
52, 4anbi12d 630 . . . 4 (𝑔 = 𝐺 → ((𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)) ↔ (𝐺r𝐹 ∧ (∫1𝐺) = (∫1𝐺))))
65rspcev 3611 . . 3 ((𝐺 ∈ dom ∫1 ∧ (𝐺r𝐹 ∧ (∫1𝐺) = (∫1𝐺))) → ∃𝑔 ∈ dom ∫1(𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
71, 6mpanr2 702 . 2 ((𝐺 ∈ dom ∫1𝐺r𝐹) → ∃𝑔 ∈ dom ∫1(𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
8 itg2val.1 . . 3 𝐿 = {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔r𝐹𝑥 = (∫1𝑔))}
98itg2l 25679 . 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 394   = wceq 1533  wcel 2098  {cab 2705  wrex 3067   class class class wbr 5152  dom cdm 5682  cfv 6553  r cofr 7690  cle 11287  1citg1 25564
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2699  ax-nul 5310
This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-sb 2060  df-clab 2706  df-cleq 2720  df-clel 2806  df-ne 2938  df-ral 3059  df-rex 3068  df-rab 3431  df-v 3475  df-dif 3952  df-un 3954  df-in 3956  df-ss 3966  df-nul 4327  df-if 4533  df-sn 4633  df-pr 4635  df-op 4639  df-uni 4913  df-br 5153  df-iota 6505  df-fv 6561
This theorem is referenced by:  itg2ub  25683
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