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Theorem itg2lr 23542
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 2651 . . 3 (∫1𝐺) = (∫1𝐺)
2 breq1 4688 . . . . 5 (𝑔 = 𝐺 → (𝑔𝑟𝐹𝐺𝑟𝐹))
3 fveq2 6229 . . . . . 6 (𝑔 = 𝐺 → (∫1𝑔) = (∫1𝐺))
43eqeq2d 2661 . . . . 5 (𝑔 = 𝐺 → ((∫1𝐺) = (∫1𝑔) ↔ (∫1𝐺) = (∫1𝐺)))
52, 4anbi12d 747 . . . 4 (𝑔 = 𝐺 → ((𝑔𝑟𝐹 ∧ (∫1𝐺) = (∫1𝑔)) ↔ (𝐺𝑟𝐹 ∧ (∫1𝐺) = (∫1𝐺))))
65rspcev 3340 . . 3 ((𝐺 ∈ dom ∫1 ∧ (𝐺𝑟𝐹 ∧ (∫1𝐺) = (∫1𝐺))) → ∃𝑔 ∈ dom ∫1(𝑔𝑟𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
71, 6mpanr2 720 . 2 ((𝐺 ∈ dom ∫1𝐺𝑟𝐹) → ∃𝑔 ∈ dom ∫1(𝑔𝑟𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
8 itg2val.1 . . 3 𝐿 = {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔𝑟𝐹𝑥 = (∫1𝑔))}
98itg2l 23541 . 2 ((∫1𝐺) ∈ 𝐿 ↔ ∃𝑔 ∈ dom ∫1(𝑔𝑟𝐹 ∧ (∫1𝐺) = (∫1𝑔)))
107, 9sylibr 224 1 ((𝐺 ∈ dom ∫1𝐺𝑟𝐹) → (∫1𝐺) ∈ 𝐿)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 383   = wceq 1523  wcel 2030  {cab 2637  wrex 2942   class class class wbr 4685  dom cdm 5143  cfv 5926  𝑟 cofr 6938  cle 10113  1citg1 23429
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631  ax-nul 4822
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1056  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-eu 2502  df-clab 2638  df-cleq 2644  df-clel 2647  df-nfc 2782  df-ral 2946  df-rex 2947  df-rab 2950  df-v 3233  df-sbc 3469  df-dif 3610  df-un 3612  df-in 3614  df-ss 3621  df-nul 3949  df-if 4120  df-sn 4211  df-pr 4213  df-op 4217  df-uni 4469  df-br 4686  df-iota 5889  df-fv 5934
This theorem is referenced by:  itg2ub  23545
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