Theorem divgcdnnr

Description: A positive integer divided by the gcd of it and another integer is a positive integer. (Contributed by AV, 10-Jul-2021)

	Ref	Expression
Assertion	divgcdnnr	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ ) → ( 𝐴 / ( 𝐵 gcd 𝐴 ) ) ∈ ℕ )

Proof

Step	Hyp	Ref	Expression
1		nnz	⊢ ( 𝐴 ∈ ℕ → 𝐴 ∈ ℤ )
2		gcdcom	⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → ( 𝐴 gcd 𝐵 ) = ( 𝐵 gcd 𝐴 ) )
3	1 2	sylan	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ ) → ( 𝐴 gcd 𝐵 ) = ( 𝐵 gcd 𝐴 ) )
4	3	eqcomd	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ ) → ( 𝐵 gcd 𝐴 ) = ( 𝐴 gcd 𝐵 ) )
5	4	oveq2d	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ ) → ( 𝐴 / ( 𝐵 gcd 𝐴 ) ) = ( 𝐴 / ( 𝐴 gcd 𝐵 ) ) )
6		divgcdnn	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ ) → ( 𝐴 / ( 𝐴 gcd 𝐵 ) ) ∈ ℕ )
7	5 6	eqeltrd	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ ) → ( 𝐴 / ( 𝐵 gcd 𝐴 ) ) ∈ ℕ )