dvdsexpim

Description: dvdssqim generalized to nonnegative exponents. (Contributed by Steven Nguyen, 2-Apr-2023)

		Ref	Expression
	Assertion	dvdsexpim	$⊢ (A \in ℤ \land B \in ℤ \land N \in ℕ_{0}) \to (A ∥ B \to A^{N} ∥ B^{N})$

Proof

Step	Hyp	Ref	Expression
1		divides	$⊢ (A \in ℤ \land B \in ℤ) \to (A ∥ B \leftrightarrow \exists k \in ℤ k A = B)$
2	1	3adant3	$⊢ (A \in ℤ \land B \in ℤ \land N \in ℕ_{0}) \to (A ∥ B \leftrightarrow \exists k \in ℤ k A = B)$
3		zexpcl	$⊢ (k \in ℤ \land N \in ℕ_{0}) \to k^{N} \in ℤ$
4	3	ancoms	$⊢ (N \in ℕ_{0} \land k \in ℤ) \to k^{N} \in ℤ$
5	4	adantll	$⊢ ((A \in ℤ \land N \in ℕ_{0}) \land k \in ℤ) \to k^{N} \in ℤ$
6		zexpcl	$⊢ (A \in ℤ \land N \in ℕ_{0}) \to A^{N} \in ℤ$
7	6	adantr	$⊢ ((A \in ℤ \land N \in ℕ_{0}) \land k \in ℤ) \to A^{N} \in ℤ$
8		dvdsmul2	$⊢ (k^{N} \in ℤ \land A^{N} \in ℤ) \to A^{N} ∥ k^{N} A^{N}$
9	5 7 8	syl2anc	$⊢ ((A \in ℤ \land N \in ℕ_{0}) \land k \in ℤ) \to A^{N} ∥ k^{N} A^{N}$
10		zcn	$⊢ k \in ℤ \to k \in ℂ$
11	10	adantl	$⊢ ((A \in ℤ \land N \in ℕ_{0}) \land k \in ℤ) \to k \in ℂ$
12		zcn	$⊢ A \in ℤ \to A \in ℂ$
13	12	ad2antrr	$⊢ ((A \in ℤ \land N \in ℕ_{0}) \land k \in ℤ) \to A \in ℂ$
14		simplr	$⊢ ((A \in ℤ \land N \in ℕ_{0}) \land k \in ℤ) \to N \in ℕ_{0}$
15	11 13 14	mulexpd	$⊢ ((A \in ℤ \land N \in ℕ_{0}) \land k \in ℤ) \to {k A}^{N} = k^{N} A^{N}$
16	9 15	breqtrrd	$⊢ ((A \in ℤ \land N \in ℕ_{0}) \land k \in ℤ) \to A^{N} ∥ {k A}^{N}$
17		oveq1	$⊢ k A = B \to {k A}^{N} = B^{N}$
18	17	breq2d	$⊢ k A = B \to (A^{N} ∥ {k A}^{N} \leftrightarrow A^{N} ∥ B^{N})$
19	16 18	syl5ibcom	$⊢ ((A \in ℤ \land N \in ℕ_{0}) \land k \in ℤ) \to (k A = B \to A^{N} ∥ B^{N})$
20	19	rexlimdva	$⊢ (A \in ℤ \land N \in ℕ_{0}) \to (\exists k \in ℤ k A = B \to A^{N} ∥ B^{N})$
21	20	3adant2	$⊢ (A \in ℤ \land B \in ℤ \land N \in ℕ_{0}) \to (\exists k \in ℤ k A = B \to A^{N} ∥ B^{N})$
22	2 21	sylbid	$⊢ (A \in ℤ \land B \in ℤ \land N \in ℕ_{0}) \to (A ∥ B \to A^{N} ∥ B^{N})$

Metamath Proof Explorer

Theorem dvdsexpim

Proof