Metamath Proof Explorer


Theorem ovolfsval

Description: The value of the interval length function. (Contributed by Mario Carneiro, 16-Mar-2014)

Ref Expression
Hypothesis ovolfs.1 ⊢ G = abs ∘ − ∘ F
Assertion ovolfsval ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → G ⁡ N = 2 nd ⁡ F ⁡ N − 1 st ⁡ F ⁡ N

Proof

Step Hyp Ref Expression
1 ovolfs.1 ⊢ G = abs ∘ − ∘ F
2 1 fveq1i ⊢ G ⁡ N = abs ∘ − ∘ F ⁡ N
3 fvco3 ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → abs ∘ − ∘ F ⁡ N = abs ∘ − ⁡ F ⁡ N
4 2 3 eqtrid ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → G ⁡ N = abs ∘ − ⁡ F ⁡ N
5 ffvelcdm ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → F ⁡ N ∈ ≤ ∩ ℝ 2
6 5 elin2d ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → F ⁡ N ∈ ℝ 2
7 1st2nd2 ⊢ F ⁡ N ∈ ℝ 2 → F ⁡ N = 1 st ⁡ F ⁡ N 2 nd ⁡ F ⁡ N
8 6 7 syl ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → F ⁡ N = 1 st ⁡ F ⁡ N 2 nd ⁡ F ⁡ N
9 8 fveq2d ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → abs ∘ − ⁡ F ⁡ N = abs ∘ − ⁡ 1 st ⁡ F ⁡ N 2 nd ⁡ F ⁡ N
10 df-ov ⊢ 1 st ⁡ F ⁡ N abs ∘ − 2 nd ⁡ F ⁡ N = abs ∘ − ⁡ 1 st ⁡ F ⁡ N 2 nd ⁡ F ⁡ N
11 9 10 eqtr4di ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → abs ∘ − ⁡ F ⁡ N = 1 st ⁡ F ⁡ N abs ∘ − 2 nd ⁡ F ⁡ N
12 ovolfcl ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → 1 st ⁡ F ⁡ N ∈ ℝ ∧ 2 nd ⁡ F ⁡ N ∈ ℝ ∧ 1 st ⁡ F ⁡ N ≤ 2 nd ⁡ F ⁡ N
13 12 simp1d ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → 1 st ⁡ F ⁡ N ∈ ℝ
14 13 recnd ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → 1 st ⁡ F ⁡ N ∈ ℂ
15 12 simp2d ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → 2 nd ⁡ F ⁡ N ∈ ℝ
16 15 recnd ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → 2 nd ⁡ F ⁡ N ∈ ℂ
17 eqid ⊢ abs ∘ − = abs ∘ −
18 17 cnmetdval ⊢ 1 st ⁡ F ⁡ N ∈ ℂ ∧ 2 nd ⁡ F ⁡ N ∈ ℂ → 1 st ⁡ F ⁡ N abs ∘ − 2 nd ⁡ F ⁡ N = 1 st ⁡ F ⁡ N − 2 nd ⁡ F ⁡ N
19 14 16 18 syl2anc ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → 1 st ⁡ F ⁡ N abs ∘ − 2 nd ⁡ F ⁡ N = 1 st ⁡ F ⁡ N − 2 nd ⁡ F ⁡ N
20 abssuble0 ⊢ 1 st ⁡ F ⁡ N ∈ ℝ ∧ 2 nd ⁡ F ⁡ N ∈ ℝ ∧ 1 st ⁡ F ⁡ N ≤ 2 nd ⁡ F ⁡ N → 1 st ⁡ F ⁡ N − 2 nd ⁡ F ⁡ N = 2 nd ⁡ F ⁡ N − 1 st ⁡ F ⁡ N
21 12 20 syl ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → 1 st ⁡ F ⁡ N − 2 nd ⁡ F ⁡ N = 2 nd ⁡ F ⁡ N − 1 st ⁡ F ⁡ N
22 19 21 eqtrd ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → 1 st ⁡ F ⁡ N abs ∘ − 2 nd ⁡ F ⁡ N = 2 nd ⁡ F ⁡ N − 1 st ⁡ F ⁡ N
23 11 22 eqtrd ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → abs ∘ − ⁡ F ⁡ N = 2 nd ⁡ F ⁡ N − 1 st ⁡ F ⁡ N
24 4 23 eqtrd ⊢ F : ℕ ⟶ ≤ ∩ ℝ 2 ∧ N ∈ ℕ → G ⁡ N = 2 nd ⁡ F ⁡ N − 1 st ⁡ F ⁡ N