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Theorem itg2lr 23897
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 2826 . . 3 (∫1𝐺) = (∫1𝐺)
2 breq1 4877 . . . . 5 (𝑔 = 𝐺 → (𝑔𝑟𝐹𝐺𝑟𝐹))
3 fveq2 6434 . . . . . 6 (𝑔 = 𝐺 → (∫1𝑔) = (∫1𝐺))
43eqeq2d 2836 . . . . 5 (𝑔 = 𝐺 → ((∫1𝐺) = (∫1𝑔) ↔ (∫1𝐺) = (∫1𝐺)))
52, 4anbi12d 626 . . . 4 (𝑔 = 𝐺 → ((𝑔𝑟𝐹 ∧ (∫1𝐺) = (∫1𝑔)) ↔ (𝐺𝑟𝐹 ∧ (∫1𝐺) = (∫1𝐺))))
65rspcev 3527 . . 3 ((𝐺 ∈ dom ∫1 ∧ (𝐺𝑟𝐹 ∧ (∫1𝐺) = (∫1𝐺))) → ∃𝑔 ∈ dom ∫1(𝑔𝑟𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
71, 6mpanr2 697 . 2 ((𝐺 ∈ dom ∫1𝐺𝑟𝐹) → ∃𝑔 ∈ dom ∫1(𝑔𝑟𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
8 itg2val.1 . . 3 𝐿 = {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔𝑟𝐹𝑥 = (∫1𝑔))}
98itg2l 23896 . 2 ((∫1𝐺) ∈ 𝐿 ↔ ∃𝑔 ∈ dom ∫1(𝑔𝑟𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
107, 9sylibr 226 1 ((𝐺 ∈ dom ∫1𝐺𝑟𝐹) → (∫1𝐺) ∈ 𝐿)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 386   = wceq 1658  wcel 2166  {cab 2812  wrex 3119   class class class wbr 4874  dom cdm 5343  cfv 6124  𝑟 cofr 7157  cle 10393  1citg1 23782
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1896  ax-4 1910  ax-5 2011  ax-6 2077  ax-7 2114  ax-9 2175  ax-10 2194  ax-11 2209  ax-12 2222  ax-13 2391  ax-ext 2804  ax-nul 5014
This theorem depends on definitions:  df-bi 199  df-an 387  df-or 881  df-3an 1115  df-tru 1662  df-ex 1881  df-nf 1885  df-sb 2070  df-mo 2606  df-eu 2641  df-clab 2813  df-cleq 2819  df-clel 2822  df-nfc 2959  df-ral 3123  df-rex 3124  df-rab 3127  df-v 3417  df-sbc 3664  df-dif 3802  df-un 3804  df-in 3806  df-ss 3813  df-nul 4146  df-if 4308  df-sn 4399  df-pr 4401  df-op 4405  df-uni 4660  df-br 4875  df-iota 6087  df-fv 6132
This theorem is referenced by:  itg2ub  23900
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