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Theorem itg2lr 26084
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(𝑔 ∘r ≤ 𝐹 ∧ 𝑥 = (∫1‘𝑔))}
Assertion
Ref Expression
itg2lr ((𝐺 ∈ dom ∫1 ∧ 𝐺 ∘r ≤ 𝐹) → (∫1‘𝐺) ∈ 𝐿)
Distinct variable groups:   𝑥,𝑔,𝐹   𝑔,𝐺,𝑥
Allowed substitution hints:   𝐿(𝑥, 𝑔)

Proof of Theorem itg2lr
StepHypRef Expression
1 eqid 2763 . . 3 (∫1‘𝐺) = (∫1‘𝐺)
2 breq1 5109 . . . . 5 (𝑔 = 𝐺 → (𝑔 ∘r ≤ 𝐹 ↔ 𝐺 ∘r ≤ 𝐹))
3 fveq2 6911 . . . . . 6 (𝑔 = 𝐺 → (∫1‘𝑔) = (∫1‘𝐺))
43eqeq2d 2774 . . . . 5 (𝑔 = 𝐺 → ((∫1‘𝐺) = (∫1‘𝑔) ↔ (∫1‘𝐺) = (∫1‘𝐺)))
52, 4anbi12d 644 . . . 4 (𝑔 = 𝐺 → ((𝑔 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝑔)) ↔ (𝐺 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝐺))))
65rspcev 3580 . . 3 ((𝐺 ∈ dom ∫1 ∧ (𝐺 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝐺))) → ∃𝑔 ∈ dom ∫1(𝑔 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝑔)))
71, 6mpanr2 717 . 2 ((𝐺 ∈ dom ∫1 ∧ 𝐺 ∘r ≤ 𝐹) → ∃𝑔 ∈ dom ∫1(𝑔 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝑔)))
8 itg2val.1 . . 3 𝐿 = {𝑥 ∣ ∃𝑔 ∈ dom ∫1(𝑔 ∘r ≤ 𝐹 ∧ 𝑥 = (∫1‘𝑔))}
98itg2l 26083 . 2 ((∫1‘𝐺) ∈ 𝐿 ↔ ∃𝑔 ∈ dom ∫1(𝑔 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝑔)))
107, 9sylibr 237 1 ((𝐺 ∈ dom ∫1 ∧ 𝐺 ∘r ≤ 𝐹) → (∫1‘𝐺) ∈ 𝐿)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2147  {cab 2741  ∃wrex 3090   class class class wbr 5106  dom cdm 5656  ‘cfv 6561   ∘r cofr 7719   ≤ cle 11374  ∫1citg1 25969
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-nul 5264
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-br 5107  df-iota 6515  df-fv 6569
This theorem is used by:  itg2ub  26087
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