Theorem itg2lr 25964

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

Step	Hyp	Ref	Expression
1		eqid 2762	. . 3 ⊢ (∫1‘𝐺) = (∫1‘𝐺)
2		breq1 5110	. . . . 5 ⊢ (𝑔 = 𝐺 → (𝑔 ∘r ≤ 𝐹 ↔ 𝐺 ∘r ≤ 𝐹))
3		fveq2 6882	. . . . . 6 ⊢ (𝑔 = 𝐺 → (∫1‘𝑔) = (∫1‘𝐺))
4	3	eqeq2d 2773	. . . . 5 ⊢ (𝑔 = 𝐺 → ((∫1‘𝐺) = (∫1‘𝑔) ↔ (∫1‘𝐺) = (∫1‘𝐺)))
5	2, 4	anbi12d 644	. . . 4 ⊢ (𝑔 = 𝐺 → ((𝑔 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝑔)) ↔ (𝐺 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝐺))))
6	5	rspcev 3579	. . 3 ⊢ ((𝐺 ∈ dom ∫1 ∧ (𝐺 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝐺))) → ∃𝑔 ∈ dom ∫1(𝑔 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝑔)))
7	1, 6	mpanr2 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))}
9	8	itg2l 25963	. 2 ⊢ ((∫1‘𝐺) ∈ 𝐿 ↔ ∃𝑔 ∈ dom ∫1(𝑔 ∘r ≤ 𝐹 ∧ (∫1‘𝐺) = (∫1‘𝑔)))
10	7, 9	sylibr 237	1 ⊢ ((𝐺 ∈ dom ∫1 ∧ 𝐺 ∘r ≤ 𝐹) → (∫1‘𝐺) ∈ 𝐿)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {cab 2740  ∃wrex 3088   class class class wbr 5107  dom cdm 5659  ‘cfv 6537   ∘_r cofr 7681   ≤ cle 11272  ∫₁citg1 25849

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 2734  ax-nul 5267

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 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-iota 6493  df-fv 6545

This theorem is used by:  itg2ub  25967