Metamath Proof Explorer


Theorem absexpz

Description: Absolute value of integer exponentiation. (Contributed by Mario Carneiro, 6-Apr-2015)

Ref Expression
Assertion absexpz ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℤ → A N = A N

Proof

Step Hyp Ref Expression
1 elznn0nn ⊢ N ∈ ℤ ↔ N ∈ ℕ 0 ∨ N ∈ ℝ ∧ − N ∈ ℕ
2 absexp ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N = A N
3 2 ex ⊢ A ∈ ℂ → N ∈ ℕ 0 → A N = A N
4 3 adantr ⊢ A ∈ ℂ ∧ A ≠ 0 → N ∈ ℕ 0 → A N = A N
5 1cnd ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → 1 ∈ ℂ
6 simpll ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A ∈ ℂ
7 nnnn0 ⊢ − N ∈ ℕ → − N ∈ ℕ 0
8 7 ad2antll ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → − N ∈ ℕ 0
9 6 8 expcld ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A − N ∈ ℂ
10 simplr ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A ≠ 0
11 nnz ⊢ − N ∈ ℕ → − N ∈ ℤ
12 11 ad2antll ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → − N ∈ ℤ
13 6 10 12 expne0d ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A − N ≠ 0
14 absdiv ⊢ 1 ∈ ℂ ∧ A − N ∈ ℂ ∧ A − N ≠ 0 → 1 A − N = 1 A − N
15 5 9 13 14 syl3anc ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → 1 A − N = 1 A − N
16 abs1 ⊢ 1 = 1
17 16 oveq1i ⊢ 1 A − N = 1 A − N
18 absexp ⊢ A ∈ ℂ ∧ − N ∈ ℕ 0 → A − N = A − N
19 6 8 18 syl2anc ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A − N = A − N
20 19 oveq2d ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → 1 A − N = 1 A − N
21 17 20 eqtrid ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → 1 A − N = 1 A − N
22 15 21 eqtrd ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → 1 A − N = 1 A − N
23 simprl ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → N ∈ ℝ
24 23 recnd ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → N ∈ ℂ
25 expneg2 ⊢ A ∈ ℂ ∧ N ∈ ℂ ∧ − N ∈ ℕ 0 → A N = 1 A − N
26 6 24 8 25 syl3anc ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A N = 1 A − N
27 26 fveq2d ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A N = 1 A − N
28 abscl ⊢ A ∈ ℂ → A ∈ ℝ
29 28 ad2antrr ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A ∈ ℝ
30 29 recnd ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A ∈ ℂ
31 expneg2 ⊢ A ∈ ℂ ∧ N ∈ ℂ ∧ − N ∈ ℕ 0 → A N = 1 A − N
32 30 24 8 31 syl3anc ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A N = 1 A − N
33 22 27 32 3eqtr4d ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℝ ∧ − N ∈ ℕ → A N = A N
34 33 ex ⊢ A ∈ ℂ ∧ A ≠ 0 → N ∈ ℝ ∧ − N ∈ ℕ → A N = A N
35 4 34 jaod ⊢ A ∈ ℂ ∧ A ≠ 0 → N ∈ ℕ 0 ∨ N ∈ ℝ ∧ − N ∈ ℕ → A N = A N
36 35 3impia ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℕ 0 ∨ N ∈ ℝ ∧ − N ∈ ℕ → A N = A N
37 1 36 syl3an3b ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ N ∈ ℤ → A N = A N