Theorem itg2lr 25926

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 2766	. . 3 ⊢ (∫1‘𝐺) = (∫1‘𝐺)
2		breq1 5117	. . . . 5 ⊢ (𝑔 = 𝐺 → (𝑔 ∘r ≤ 𝐹 ↔ 𝐺 ∘r ≤ 𝐹))
3		fveq2 6888	. . . . . 6 ⊢ (𝑔 = 𝐺 → (∫1‘𝑔) = (∫1‘𝐺))
4	3	eqeq2d 2777	. . . . 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 3584	. . 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 25925	. 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 2146  {cab 2744  ∃wrex 3092   class class class wbr 5114  dom cdm 5666  ‘cfv 6543   ∘_r cofr 7686   ≤ cle 11262  ∫₁citg1 25811

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 2148  ax-9 2156  ax-ext 2738  ax-nul 5274

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 2745  df-cleq 2758  df-clel 2841  df-ne 2962  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-uni 4878  df-br 5115  df-iota 6499  df-fv 6551

This theorem is used by:  itg2ub  25929