Metamath Proof Explorer


Theorem itg2l

Description: Elementhood in the set L of lower sums of the integral. (Contributed by Mario Carneiro, 28-Jun-2014)

Ref Expression
Hypothesis itg2val.1 ⊢ L = x | ∃ g ∈ dom ⁡ ∫ 1 g ≤ f F ∧ x = ∫ 1 ⁡ g
Assertion itg2l ⊢ A ∈ L ↔ ∃ g ∈ dom ⁡ ∫ 1 g ≤ f F ∧ A = ∫ 1 ⁡ g

Proof

Step Hyp Ref Expression
1 itg2val.1 ⊢ L = x | ∃ g ∈ dom ⁡ ∫ 1 g ≤ f F ∧ x = ∫ 1 ⁡ g
2 1 eleq2i ⊢ A ∈ L ↔ A ∈ x | ∃ g ∈ dom ⁡ ∫ 1 g ≤ f F ∧ x = ∫ 1 ⁡ g
3 simpr ⊢ g ≤ f F ∧ A = ∫ 1 ⁡ g → A = ∫ 1 ⁡ g
4 fvex ⊢ ∫ 1 ⁡ g ∈ V
5 3 4 eqeltrdi ⊢ g ≤ f F ∧ A = ∫ 1 ⁡ g → A ∈ V
6 5 rexlimivw ⊢ ∃ g ∈ dom ⁡ ∫ 1 g ≤ f F ∧ A = ∫ 1 ⁡ g → A ∈ V
7 eqeq1 ⊢ x = A → x = ∫ 1 ⁡ g ↔ A = ∫ 1 ⁡ g
8 7 anbi2d ⊢ x = A → g ≤ f F ∧ x = ∫ 1 ⁡ g ↔ g ≤ f F ∧ A = ∫ 1 ⁡ g
9 8 rexbidv ⊢ x = A → ∃ g ∈ dom ⁡ ∫ 1 g ≤ f F ∧ x = ∫ 1 ⁡ g ↔ ∃ g ∈ dom ⁡ ∫ 1 g ≤ f F ∧ A = ∫ 1 ⁡ g
10 6 9 elab3 ⊢ A ∈ x | ∃ g ∈ dom ⁡ ∫ 1 g ≤ f F ∧ x = ∫ 1 ⁡ g ↔ ∃ g ∈ dom ⁡ ∫ 1 g ≤ f F ∧ A = ∫ 1 ⁡ g
11 2 10 bitri ⊢ A ∈ L ↔ ∃ g ∈ dom ⁡ ∫ 1 g ≤ f F ∧ A = ∫ 1 ⁡ g