Metamath Proof Explorer


Theorem risefacp1

Description: The value of the rising factorial at a successor. (Contributed by Scott Fenton, 5-Jan-2018)

Ref Expression
Assertion risefacp1 ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N + 1 ‾ = A N ‾ ⁢ A + N

Proof

Step Hyp Ref Expression
1 nn0cn ⊢ N ∈ ℕ 0 → N ∈ ℂ
2 1 adantl ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → N ∈ ℂ
3 1cnd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → 1 ∈ ℂ
4 2 3 pncand ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → N + 1 - 1 = N
5 4 oveq2d ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → 0 … N + 1 - 1 = 0 … N
6 5 prodeq1d ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → ∏ k = 0 N + 1 - 1 A + k = ∏ k = 0 N A + k
7 elnn0uz ⊢ N ∈ ℕ 0 ↔ N ∈ ℤ ≥ 0
8 7 bilani ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → N ∈ ℤ ≥ 0
9 elfznn0 ⊢ k ∈ 0 … N → k ∈ ℕ 0
10 9 nn0cnd ⊢ k ∈ 0 … N → k ∈ ℂ
11 addcl ⊢ A ∈ ℂ ∧ k ∈ ℂ → A + k ∈ ℂ
12 10 11 sylan2 ⊢ A ∈ ℂ ∧ k ∈ 0 … N → A + k ∈ ℂ
13 12 adantlr ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ k ∈ 0 … N → A + k ∈ ℂ
14 oveq2 ⊢ k = N → A + k = A + N
15 8 13 14 fprodm1 ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → ∏ k = 0 N A + k = ∏ k = 0 N − 1 A + k ⁢ A + N
16 6 15 eqtrd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → ∏ k = 0 N + 1 - 1 A + k = ∏ k = 0 N − 1 A + k ⁢ A + N
17 peano2nn0 ⊢ N ∈ ℕ 0 → N + 1 ∈ ℕ 0
18 risefacval ⊢ A ∈ ℂ ∧ N + 1 ∈ ℕ 0 → A N + 1 ‾ = ∏ k = 0 N + 1 - 1 A + k
19 17 18 sylan2 ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N + 1 ‾ = ∏ k = 0 N + 1 - 1 A + k
20 risefacval ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N ‾ = ∏ k = 0 N − 1 A + k
21 20 oveq1d ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N ‾ ⁢ A + N = ∏ k = 0 N − 1 A + k ⁢ A + N
22 16 19 21 3eqtr4d ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N + 1 ‾ = A N ‾ ⁢ A + N