Metamath Proof Explorer


Theorem ruclem4

Description: Lemma for ruc . Initial value of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014)

Ref Expression
Hypotheses ruc.1 ⊢ φ → F : ℕ ⟶ ℝ
ruc.2 ⊢ φ → D = x ∈ ℝ 2 , y ∈ ℝ ⟼ ⦋ 1 st ⁡ x + 2 nd ⁡ x 2 / m⦌ if m < y 1 st ⁡ x m m + 2 nd ⁡ x 2 2 nd ⁡ x
ruc.4 ⊢ C = 0 0 1 ∪ F
ruc.5 ⊢ G = seq 0 D C
Assertion ruclem4 ⊢ φ → G ⁡ 0 = 0 1

Proof

Step Hyp Ref Expression
1 ruc.1 ⊢ φ → F : ℕ ⟶ ℝ
2 ruc.2 ⊢ φ → D = x ∈ ℝ 2 , y ∈ ℝ ⟼ ⦋ 1 st ⁡ x + 2 nd ⁡ x 2 / m⦌ if m < y 1 st ⁡ x m m + 2 nd ⁡ x 2 2 nd ⁡ x
3 ruc.4 ⊢ C = 0 0 1 ∪ F
4 ruc.5 ⊢ G = seq 0 D C
5 4 fveq1i ⊢ G ⁡ 0 = seq 0 D C ⁡ 0
6 0z ⊢ 0 ∈ ℤ
7 ffn ⊢ F : ℕ ⟶ ℝ → F Fn ℕ
8 fnresdm ⊢ F Fn ℕ → F ↾ ℕ = F
9 1 7 8 3syl ⊢ φ → F ↾ ℕ = F
10 dfn2 ⊢ ℕ = ℕ 0 ∖ 0
11 10 reseq2i ⊢ F ↾ ℕ = F ↾ ℕ 0 ∖ 0
12 9 11 eqtr3di ⊢ φ → F = F ↾ ℕ 0 ∖ 0
13 12 uneq2d ⊢ φ → 0 0 1 ∪ F = 0 0 1 ∪ F ↾ ℕ 0 ∖ 0
14 3 13 eqtrid ⊢ φ → C = 0 0 1 ∪ F ↾ ℕ 0 ∖ 0
15 14 fveq1d ⊢ φ → C ⁡ 0 = 0 0 1 ∪ F ↾ ℕ 0 ∖ 0 ⁡ 0
16 c0ex ⊢ 0 ∈ V
17 16 a1i ⊢ ⊤ → 0 ∈ V
18 opex ⊢ 0 1 ∈ V
19 18 a1i ⊢ ⊤ → 0 1 ∈ V
20 eqid ⊢ 0 0 1 ∪ F ↾ ℕ 0 ∖ 0 = 0 0 1 ∪ F ↾ ℕ 0 ∖ 0
21 17 19 20 fvsnun1 ⊢ ⊤ → 0 0 1 ∪ F ↾ ℕ 0 ∖ 0 ⁡ 0 = 0 1
22 21 mptru ⊢ 0 0 1 ∪ F ↾ ℕ 0 ∖ 0 ⁡ 0 = 0 1
23 15 22 eqtrdi ⊢ φ → C ⁡ 0 = 0 1
24 6 23 seq1i ⊢ φ → seq 0 D C ⁡ 0 = 0 1
25 5 24 eqtrid ⊢ φ → G ⁡ 0 = 0 1