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Theorem itg2lr 24323
 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 2819 . . 3 (∫1𝐺) = (∫1𝐺)
2 breq1 5060 . . . . 5 (𝑔 = 𝐺 → (𝑔r𝐹𝐺r𝐹))
3 fveq2 6663 . . . . . 6 (𝑔 = 𝐺 → (∫1𝑔) = (∫1𝐺))
43eqeq2d 2830 . . . . 5 (𝑔 = 𝐺 → ((∫1𝐺) = (∫1𝑔) ↔ (∫1𝐺) = (∫1𝐺)))
52, 4anbi12d 632 . . . 4 (𝑔 = 𝐺 → ((𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)) ↔ (𝐺r𝐹 ∧ (∫1𝐺) = (∫1𝐺))))
65rspcev 3621 . . 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 24322 . 2 ((∫1𝐺) ∈ 𝐿 ↔ ∃𝑔 ∈ dom ∫1(𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
107, 9sylibr 236 1 ((𝐺 ∈ dom ∫1𝐺r𝐹) → (∫1𝐺) ∈ 𝐿)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ wa 398   = wceq 1530   ∈ wcel 2107  {cab 2797  ∃wrex 3137   class class class wbr 5057  dom cdm 5548  ‘cfv 6348   ∘r cofr 7400   ≤ cle 10668  ∫1citg1 24208 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 1904  ax-6 1963  ax-7 2008  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2153  ax-12 2169  ax-ext 2791  ax-nul 5201 This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1083  df-tru 1533  df-ex 1774  df-nf 1778  df-sb 2063  df-mo 2616  df-eu 2648  df-clab 2798  df-cleq 2812  df-clel 2891  df-nfc 2961  df-ral 3141  df-rex 3142  df-rab 3145  df-v 3495  df-sbc 3771  df-dif 3937  df-un 3939  df-in 3941  df-ss 3950  df-nul 4290  df-if 4466  df-sn 4560  df-pr 4562  df-op 4566  df-uni 4831  df-br 5058  df-iota 6307  df-fv 6356 This theorem is referenced by:  itg2ub  24326
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