Metamath Proof Explorer


Theorem ef0lem

Description: The series defining the exponential function converges in the (trivial) case of a zero argument. (Contributed by Steve Rodriguez, 7-Jun-2006) (Revised by Mario Carneiro, 28-Apr-2014)

Ref Expression
Hypothesis eftval.1 ⊢ F = n ∈ ℕ 0 ⟼ A n n !
Assertion ef0lem ⊢ A = 0 → seq 0 + F ⇝ 1

Proof

Step Hyp Ref Expression
1 eftval.1 ⊢ F = n ∈ ℕ 0 ⟼ A n n !
2 simpr ⊢ A = 0 ∧ k ∈ ℤ ≥ 0 → k ∈ ℤ ≥ 0
3 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
4 2 3 eleqtrrdi ⊢ A = 0 ∧ k ∈ ℤ ≥ 0 → k ∈ ℕ 0
5 elnn0 ⊢ k ∈ ℕ 0 ↔ k ∈ ℕ ∨ k = 0
6 4 5 sylib ⊢ A = 0 ∧ k ∈ ℤ ≥ 0 → k ∈ ℕ ∨ k = 0
7 nnnn0 ⊢ k ∈ ℕ → k ∈ ℕ 0
8 7 adantl ⊢ A = 0 ∧ k ∈ ℕ → k ∈ ℕ 0
9 1 eftval ⊢ k ∈ ℕ 0 → F ⁡ k = A k k !
10 8 9 syl ⊢ A = 0 ∧ k ∈ ℕ → F ⁡ k = A k k !
11 oveq1 ⊢ A = 0 → A k = 0 k
12 0exp ⊢ k ∈ ℕ → 0 k = 0
13 11 12 sylan9eq ⊢ A = 0 ∧ k ∈ ℕ → A k = 0
14 13 oveq1d ⊢ A = 0 ∧ k ∈ ℕ → A k k ! = 0 k !
15 faccl ⊢ k ∈ ℕ 0 → k ! ∈ ℕ
16 nncn ⊢ k ! ∈ ℕ → k ! ∈ ℂ
17 nnne0 ⊢ k ! ∈ ℕ → k ! ≠ 0
18 16 17 div0d ⊢ k ! ∈ ℕ → 0 k ! = 0
19 8 15 18 3syl ⊢ A = 0 ∧ k ∈ ℕ → 0 k ! = 0
20 10 14 19 3eqtrd ⊢ A = 0 ∧ k ∈ ℕ → F ⁡ k = 0
21 nnne0 ⊢ k ∈ ℕ → k ≠ 0
22 velsn ⊢ k ∈ 0 ↔ k = 0
23 22 necon3bbii ⊢ ¬ k ∈ 0 ↔ k ≠ 0
24 21 23 sylibr ⊢ k ∈ ℕ → ¬ k ∈ 0
25 24 adantl ⊢ A = 0 ∧ k ∈ ℕ → ¬ k ∈ 0
26 25 iffalsed ⊢ A = 0 ∧ k ∈ ℕ → if k ∈ 0 1 0 = 0
27 20 26 eqtr4d ⊢ A = 0 ∧ k ∈ ℕ → F ⁡ k = if k ∈ 0 1 0
28 fveq2 ⊢ k = 0 → F ⁡ k = F ⁡ 0
29 oveq1 ⊢ A = 0 → A 0 = 0 0
30 0exp0e1 ⊢ 0 0 = 1
31 29 30 eqtrdi ⊢ A = 0 → A 0 = 1
32 31 oveq1d ⊢ A = 0 → A 0 0 ! = 1 0 !
33 0nn0 ⊢ 0 ∈ ℕ 0
34 1 eftval ⊢ 0 ∈ ℕ 0 → F ⁡ 0 = A 0 0 !
35 33 34 ax-mp ⊢ F ⁡ 0 = A 0 0 !
36 fac0 ⊢ 0 ! = 1
37 36 oveq2i ⊢ 1 0 ! = 1 1
38 1div1e1 ⊢ 1 1 = 1
39 37 38 eqtr2i ⊢ 1 = 1 0 !
40 32 35 39 3eqtr4g ⊢ A = 0 → F ⁡ 0 = 1
41 28 40 sylan9eqr ⊢ A = 0 ∧ k = 0 → F ⁡ k = 1
42 simpr ⊢ A = 0 ∧ k = 0 → k = 0
43 42 22 sylibr ⊢ A = 0 ∧ k = 0 → k ∈ 0
44 43 iftrued ⊢ A = 0 ∧ k = 0 → if k ∈ 0 1 0 = 1
45 41 44 eqtr4d ⊢ A = 0 ∧ k = 0 → F ⁡ k = if k ∈ 0 1 0
46 27 45 jaodan ⊢ A = 0 ∧ k ∈ ℕ ∨ k = 0 → F ⁡ k = if k ∈ 0 1 0
47 6 46 syldan ⊢ A = 0 ∧ k ∈ ℤ ≥ 0 → F ⁡ k = if k ∈ 0 1 0
48 33 3 eleqtri ⊢ 0 ∈ ℤ ≥ 0
49 48 a1i ⊢ A = 0 → 0 ∈ ℤ ≥ 0
50 1cnd ⊢ A = 0 ∧ k ∈ 0 → 1 ∈ ℂ
51 fz0sn ⊢ 0 … 0 = 0
52 51 eqimss2i ⊢ 0 ⊆ 0 … 0
53 52 a1i ⊢ A = 0 → 0 ⊆ 0 … 0
54 47 49 50 53 fsumcvg2 ⊢ A = 0 → seq 0 + F ⇝ seq 0 + F ⁡ 0
55 0z ⊢ 0 ∈ ℤ
56 55 40 seq1i ⊢ A = 0 → seq 0 + F ⁡ 0 = 1
57 54 56 breqtrd ⊢ A = 0 → seq 0 + F ⇝ 1