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Theorem itg2lr 26031
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 2761 . . 3 (∫1‘𝐺) = (∫1‘𝐺)
2 breq1 5106 . . . . 5 (𝑔 = 𝐺 → (𝑔 ∘r ≤ 𝐹 ↔ 𝐺 ∘r ≤ 𝐹))
3 fveq2 6877 . . . . . 6 (𝑔 = 𝐺 → (∫1‘𝑔) = (∫1‘𝐺))
43eqeq2d 2772 . . . . 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 3577 . . 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 26030 . 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 2145  {cab 2739  ∃wrex 3087   class class class wbr 5103  dom cdm 5651  ‘cfv 6531   ∘r cofr 7681   ≤ cle 11325  ∫1citg1 25916
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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733  ax-nul 5260
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 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-iota 6487  df-fv 6539
This theorem is used by:  itg2ub  26034
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