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Theorem itg2lr 24331
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 2821 . . 3 (∫1𝐺) = (∫1𝐺)
2 breq1 5069 . . . . 5 (𝑔 = 𝐺 → (𝑔r𝐹𝐺r𝐹))
3 fveq2 6670 . . . . . 6 (𝑔 = 𝐺 → (∫1𝑔) = (∫1𝐺))
43eqeq2d 2832 . . . . 5 (𝑔 = 𝐺 → ((∫1𝐺) = (∫1𝑔) ↔ (∫1𝐺) = (∫1𝐺)))
52, 4anbi12d 632 . . . 4 (𝑔 = 𝐺 → ((𝑔r𝐹 ∧ (∫1𝐺) = (∫1𝑔)) ↔ (𝐺r𝐹 ∧ (∫1𝐺) = (∫1𝐺))))
65rspcev 3623 . . 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 24330 . 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 1537  wcel 2114  {cab 2799  wrex 3139   class class class wbr 5066  dom cdm 5555  cfv 6355  r cofr 7408  cle 10676  1citg1 24216
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2793  ax-nul 5210
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-mo 2622  df-eu 2654  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-rex 3144  df-rab 3147  df-v 3496  df-sbc 3773  df-dif 3939  df-un 3941  df-in 3943  df-ss 3952  df-nul 4292  df-if 4468  df-sn 4568  df-pr 4570  df-op 4574  df-uni 4839  df-br 5067  df-iota 6314  df-fv 6363
This theorem is referenced by:  itg2ub  24334
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