Metamath Proof Explorer


Theorem absexp

Description: Absolute value of positive integer exponentiation. (Contributed by NM, 5-Jan-2006)

Ref Expression
Assertion absexp ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N = A N

Proof

Step Hyp Ref Expression
1 oveq2 ⊢ j = 0 → A j = A 0
2 1 fveq2d ⊢ j = 0 → A j = A 0
3 oveq2 ⊢ j = 0 → A j = A 0
4 2 3 eqeq12d ⊢ j = 0 → A j = A j ↔ A 0 = A 0
5 oveq2 ⊢ j = k → A j = A k
6 5 fveq2d ⊢ j = k → A j = A k
7 oveq2 ⊢ j = k → A j = A k
8 6 7 eqeq12d ⊢ j = k → A j = A j ↔ A k = A k
9 oveq2 ⊢ j = k + 1 → A j = A k + 1
10 9 fveq2d ⊢ j = k + 1 → A j = A k + 1
11 oveq2 ⊢ j = k + 1 → A j = A k + 1
12 10 11 eqeq12d ⊢ j = k + 1 → A j = A j ↔ A k + 1 = A k + 1
13 oveq2 ⊢ j = N → A j = A N
14 13 fveq2d ⊢ j = N → A j = A N
15 oveq2 ⊢ j = N → A j = A N
16 14 15 eqeq12d ⊢ j = N → A j = A j ↔ A N = A N
17 abs1 ⊢ 1 = 1
18 exp0 ⊢ A ∈ ℂ → A 0 = 1
19 18 fveq2d ⊢ A ∈ ℂ → A 0 = 1
20 abscl ⊢ A ∈ ℂ → A ∈ ℝ
21 20 recnd ⊢ A ∈ ℂ → A ∈ ℂ
22 21 exp0d ⊢ A ∈ ℂ → A 0 = 1
23 17 19 22 3eqtr4a ⊢ A ∈ ℂ → A 0 = A 0
24 oveq1 ⊢ A k = A k → A k ⁢ A = A k ⁢ A
25 24 adantl ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 ∧ A k = A k → A k ⁢ A = A k ⁢ A
26 expp1 ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 → A k + 1 = A k ⁢ A
27 26 fveq2d ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 → A k + 1 = A k ⁢ A
28 expcl ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 → A k ∈ ℂ
29 simpl ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 → A ∈ ℂ
30 absmul ⊢ A k ∈ ℂ ∧ A ∈ ℂ → A k ⁢ A = A k ⁢ A
31 28 29 30 syl2anc ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 → A k ⁢ A = A k ⁢ A
32 27 31 eqtrd ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 → A k + 1 = A k ⁢ A
33 32 adantr ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 ∧ A k = A k → A k + 1 = A k ⁢ A
34 expp1 ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 → A k + 1 = A k ⁢ A
35 21 34 sylan ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 → A k + 1 = A k ⁢ A
36 35 adantr ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 ∧ A k = A k → A k + 1 = A k ⁢ A
37 25 33 36 3eqtr4d ⊢ A ∈ ℂ ∧ k ∈ ℕ 0 ∧ A k = A k → A k + 1 = A k + 1
38 4 8 12 16 23 37 nn0indd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N = A N