Metamath Proof Explorer


Theorem fwddifn0

Description: The value of the n-iterated forward difference operator at zero is just the function value. (Contributed by Scott Fenton, 28-May-2020)

Ref Expression
Hypotheses fwddifn0.1 ⊢ φ → A ⊆ ℂ
fwddifn0.2 ⊢ φ → F : A ⟶ ℂ
fwddifn0.3 ⊢ φ → X ∈ A
Assertion fwddifn0 ⊢ φ → 0 △n F ⁡ X = F ⁡ X

Proof

Step Hyp Ref Expression
1 fwddifn0.1 ⊢ φ → A ⊆ ℂ
2 fwddifn0.2 ⊢ φ → F : A ⟶ ℂ
3 fwddifn0.3 ⊢ φ → X ∈ A
4 0nn0 ⊢ 0 ∈ ℕ 0
5 4 a1i ⊢ φ → 0 ∈ ℕ 0
6 1 3 sseldd ⊢ φ → X ∈ ℂ
7 0z ⊢ 0 ∈ ℤ
8 fzsn ⊢ 0 ∈ ℤ → 0 … 0 = 0
9 7 8 ax-mp ⊢ 0 … 0 = 0
10 9 eleq2i ⊢ k ∈ 0 … 0 ↔ k ∈ 0
11 velsn ⊢ k ∈ 0 ↔ k = 0
12 10 11 bitri ⊢ k ∈ 0 … 0 ↔ k = 0
13 oveq2 ⊢ k = 0 → X + k = X + 0
14 13 adantl ⊢ φ ∧ k = 0 → X + k = X + 0
15 6 addridd ⊢ φ → X + 0 = X
16 15 3 eqeltrd ⊢ φ → X + 0 ∈ A
17 16 adantr ⊢ φ ∧ k = 0 → X + 0 ∈ A
18 14 17 eqeltrd ⊢ φ ∧ k = 0 → X + k ∈ A
19 12 18 sylan2b ⊢ φ ∧ k ∈ 0 … 0 → X + k ∈ A
20 5 1 2 6 19 fwddifnval ⊢ φ → 0 △n F ⁡ X = ∑ k = 0 0 ( 0 k) ⁢ − 1 0 − k ⁢ F ⁡ X + k
21 15 fveq2d ⊢ φ → F ⁡ X + 0 = F ⁡ X
22 21 oveq2d ⊢ φ → 1 ⁢ F ⁡ X + 0 = 1 ⁢ F ⁡ X
23 2 3 ffvelcdmd ⊢ φ → F ⁡ X ∈ ℂ
24 23 mullidd ⊢ φ → 1 ⁢ F ⁡ X = F ⁡ X
25 22 24 eqtrd ⊢ φ → 1 ⁢ F ⁡ X + 0 = F ⁡ X
26 25 oveq2d ⊢ φ → 1 ⁢ 1 ⁢ F ⁡ X + 0 = 1 ⁢ F ⁡ X
27 26 24 eqtrd ⊢ φ → 1 ⁢ 1 ⁢ F ⁡ X + 0 = F ⁡ X
28 27 23 eqeltrd ⊢ φ → 1 ⁢ 1 ⁢ F ⁡ X + 0 ∈ ℂ
29 oveq2 ⊢ k = 0 → ( 0 k) = ( 0 0 )
30 bcnn ⊢ 0 ∈ ℕ 0 → ( 0 0 ) = 1
31 4 30 ax-mp ⊢ ( 0 0 ) = 1
32 29 31 eqtrdi ⊢ k = 0 → ( 0 k) = 1
33 oveq2 ⊢ k = 0 → 0 − k = 0 − 0
34 0m0e0 ⊢ 0 − 0 = 0
35 33 34 eqtrdi ⊢ k = 0 → 0 − k = 0
36 35 oveq2d ⊢ k = 0 → − 1 0 − k = − 1 0
37 neg1cn ⊢ − 1 ∈ ℂ
38 exp0 ⊢ − 1 ∈ ℂ → − 1 0 = 1
39 37 38 ax-mp ⊢ − 1 0 = 1
40 36 39 eqtrdi ⊢ k = 0 → − 1 0 − k = 1
41 13 fveq2d ⊢ k = 0 → F ⁡ X + k = F ⁡ X + 0
42 40 41 oveq12d ⊢ k = 0 → − 1 0 − k ⁢ F ⁡ X + k = 1 ⁢ F ⁡ X + 0
43 32 42 oveq12d ⊢ k = 0 → ( 0 k) ⁢ − 1 0 − k ⁢ F ⁡ X + k = 1 ⁢ 1 ⁢ F ⁡ X + 0
44 43 fsum1 ⊢ 0 ∈ ℤ ∧ 1 ⁢ 1 ⁢ F ⁡ X + 0 ∈ ℂ → ∑ k = 0 0 ( 0 k) ⁢ − 1 0 − k ⁢ F ⁡ X + k = 1 ⁢ 1 ⁢ F ⁡ X + 0
45 7 28 44 sylancr ⊢ φ → ∑ k = 0 0 ( 0 k) ⁢ − 1 0 − k ⁢ F ⁡ X + k = 1 ⁢ 1 ⁢ F ⁡ X + 0
46 45 27 eqtrd ⊢ φ → ∑ k = 0 0 ( 0 k) ⁢ − 1 0 − k ⁢ F ⁡ X + k = F ⁡ X
47 20 46 eqtrd ⊢ φ → 0 △n F ⁡ X = F ⁡ X