Metamath Proof Explorer


Theorem dvdsexpim

Description: If two numbers are divisible, so are their nonnegative exponents. Similar to dvdssqim for nonnegative exponents. (Contributed by Steven Nguyen, 2-Apr-2023)

Ref Expression
Assertion dvdsexpim ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ N ∈ ℕ 0 → A ∥ B → A N ∥ B N

Proof

Step Hyp Ref Expression
1 divides ⊢ A ∈ ℤ ∧ B ∈ ℤ → A ∥ B ↔ ∃ k ∈ ℤ k ⁢ A = B
2 1 3adant3 ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ N ∈ ℕ 0 → A ∥ B ↔ ∃ k ∈ ℤ k ⁢ A = B
3 zexpcl ⊢ k ∈ ℤ ∧ N ∈ ℕ 0 → k N ∈ ℤ
4 3 ancoms ⊢ N ∈ ℕ 0 ∧ k ∈ ℤ → k N ∈ ℤ
5 4 adantll ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 ∧ k ∈ ℤ → k N ∈ ℤ
6 zexpcl ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 → A N ∈ ℤ
7 6 adantr ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 ∧ k ∈ ℤ → A N ∈ ℤ
8 dvdsmul2 ⊢ k N ∈ ℤ ∧ A N ∈ ℤ → A N ∥ k N ⁢ A N
9 5 7 8 syl2anc ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 ∧ k ∈ ℤ → A N ∥ k N ⁢ A N
10 zcn ⊢ k ∈ ℤ → k ∈ ℂ
11 10 adantl ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 ∧ k ∈ ℤ → k ∈ ℂ
12 zcn ⊢ A ∈ ℤ → A ∈ ℂ
13 12 ad2antrr ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 ∧ k ∈ ℤ → A ∈ ℂ
14 simplr ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 ∧ k ∈ ℤ → N ∈ ℕ 0
15 11 13 14 mulexpd ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 ∧ k ∈ ℤ → k ⁢ A N = k N ⁢ A N
16 9 15 breqtrrd ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 ∧ k ∈ ℤ → A N ∥ k ⁢ A N
17 oveq1 ⊢ k ⁢ A = B → k ⁢ A N = B N
18 17 breq2d ⊢ k ⁢ A = B → A N ∥ k ⁢ A N ↔ A N ∥ B N
19 16 18 syl5ibcom ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 ∧ k ∈ ℤ → k ⁢ A = B → A N ∥ B N
20 19 rexlimdva ⊢ A ∈ ℤ ∧ N ∈ ℕ 0 → ∃ k ∈ ℤ k ⁢ A = B → A N ∥ B N
21 20 3adant2 ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ N ∈ ℕ 0 → ∃ k ∈ ℤ k ⁢ A = B → A N ∥ B N
22 2 21 sylbid ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ N ∈ ℕ 0 → A ∥ B → A N ∥ B N