Metamath Proof Explorer


Theorem dvnp1

Description: Successor iterated derivative. (Contributed by Stefan O'Rear, 15-Nov-2014) (Revised by Mario Carneiro, 11-Feb-2015)

Ref Expression
Assertion dvnp1 ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → S D n F ⁡ N + 1 = S D S D n F ⁡ N

Proof

Step Hyp Ref Expression
1 simp3 ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → N ∈ ℕ 0
2 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
3 1 2 eleqtrdi ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → N ∈ ℤ ≥ 0
4 seqp1 ⊢ N ∈ ℤ ≥ 0 → seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N + 1 = seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N + 1
5 3 4 syl ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N + 1 = seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N + 1
6 fvex ⊢ seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N ∈ V
7 fvex ⊢ ℕ 0 × F ⁡ N + 1 ∈ V
8 6 7 opco1i ⊢ seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N + 1 = x ∈ V ⟼ S D x ⁡ seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N
9 5 8 eqtrdi ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N + 1 = x ∈ V ⟼ S D x ⁡ seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N
10 eqid ⊢ x ∈ V ⟼ S D x = x ∈ V ⟼ S D x
11 10 dvnfval ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S → S D n F = seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F
12 11 3adant3 ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → S D n F = seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F
13 12 fveq1d ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → S D n F ⁡ N + 1 = seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N + 1
14 fvex ⊢ S D n F ⁡ N ∈ V
15 oveq2 ⊢ x = S D n F ⁡ N → S D x = S D S D n F ⁡ N
16 ovex ⊢ S D S D n F ⁡ N ∈ V
17 15 10 16 fvmpt ⊢ S D n F ⁡ N ∈ V → x ∈ V ⟼ S D x ⁡ S D n F ⁡ N = S D S D n F ⁡ N
18 14 17 ax-mp ⊢ x ∈ V ⟼ S D x ⁡ S D n F ⁡ N = S D S D n F ⁡ N
19 12 fveq1d ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → S D n F ⁡ N = seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N
20 19 fveq2d ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → x ∈ V ⟼ S D x ⁡ S D n F ⁡ N = x ∈ V ⟼ S D x ⁡ seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N
21 18 20 eqtr3id ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → S D S D n F ⁡ N = x ∈ V ⟼ S D x ⁡ seq 0 x ∈ V ⟼ S D x ∘ 1 st ℕ 0 × F ⁡ N
22 9 13 21 3eqtr4d ⊢ S ⊆ ℂ ∧ F ∈ ℂ ↑ 𝑝𝑚 S ∧ N ∈ ℕ 0 → S D n F ⁡ N + 1 = S D S D n F ⁡ N