Metamath Proof Explorer


Theorem fprodfac

Description: Factorial using product notation. (Contributed by Scott Fenton, 15-Dec-2017)

Ref Expression
Assertion fprodfac ⊢ A ∈ ℕ 0 → A ! = ∏ k = 1 A k

Proof

Step Hyp Ref Expression
1 elnn0 ⊢ A ∈ ℕ 0 ↔ A ∈ ℕ ∨ A = 0
2 facnn ⊢ A ∈ ℕ → A ! = seq 1 × I ⁡ A
3 vex ⊢ k ∈ V
4 fvi ⊢ k ∈ V → I ⁡ k = k
5 3 4 mp1i ⊢ A ∈ ℕ ∧ k ∈ 1 … A → I ⁡ k = k
6 elnnuz ⊢ A ∈ ℕ ↔ A ∈ ℤ ≥ 1
7 6 biimpi ⊢ A ∈ ℕ → A ∈ ℤ ≥ 1
8 elfznn ⊢ k ∈ 1 … A → k ∈ ℕ
9 8 nncnd ⊢ k ∈ 1 … A → k ∈ ℂ
10 9 adantl ⊢ A ∈ ℕ ∧ k ∈ 1 … A → k ∈ ℂ
11 5 7 10 fprodser ⊢ A ∈ ℕ → ∏ k = 1 A k = seq 1 × I ⁡ A
12 2 11 eqtr4d ⊢ A ∈ ℕ → A ! = ∏ k = 1 A k
13 prod0 ⊢ ∏ k ∈ ∅ k = 1
14 13 eqcomi ⊢ 1 = ∏ k ∈ ∅ k
15 fveq2 ⊢ A = 0 → A ! = 0 !
16 fac0 ⊢ 0 ! = 1
17 15 16 eqtrdi ⊢ A = 0 → A ! = 1
18 oveq2 ⊢ A = 0 → 1 … A = 1 … 0
19 fz10 ⊢ 1 … 0 = ∅
20 18 19 eqtrdi ⊢ A = 0 → 1 … A = ∅
21 20 prodeq1d ⊢ A = 0 → ∏ k = 1 A k = ∏ k ∈ ∅ k
22 14 17 21 3eqtr4a ⊢ A = 0 → A ! = ∏ k = 1 A k
23 12 22 jaoi ⊢ A ∈ ℕ ∨ A = 0 → A ! = ∏ k = 1 A k
24 1 23 sylbi ⊢ A ∈ ℕ 0 → A ! = ∏ k = 1 A k