Theorem gcdmultiplez

Description: The GCD of a multiple of an integer is the integer itself. (Contributed by Scott Fenton, 18-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014) (Proof shortened by AV, 12-Jan-2023)

	Ref	Expression
Assertion	gcdmultiplez	⊢ ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 gcd ( 𝑀 · 𝑁 ) ) = 𝑀 )

Proof

Step	Hyp	Ref	Expression
1		nncn	⊢ ( 𝑀 ∈ ℕ → 𝑀 ∈ ℂ )
2	1	adantr	⊢ ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ ) → 𝑀 ∈ ℂ )
3		zcn	⊢ ( 𝑁 ∈ ℤ → 𝑁 ∈ ℂ )
4	3	adantl	⊢ ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ ) → 𝑁 ∈ ℂ )
5	2 4	mulcomd	⊢ ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 · 𝑁 ) = ( 𝑁 · 𝑀 ) )
6	5	oveq2d	⊢ ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 gcd ( 𝑀 · 𝑁 ) ) = ( 𝑀 gcd ( 𝑁 · 𝑀 ) ) )
7		nnnn0	⊢ ( 𝑀 ∈ ℕ → 𝑀 ∈ ℕ0 )
8	7	adantr	⊢ ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ ) → 𝑀 ∈ ℕ0 )
9		simpr	⊢ ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ ) → 𝑁 ∈ ℤ )
10	8 9	gcdmultipled	⊢ ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 gcd ( 𝑁 · 𝑀 ) ) = 𝑀 )
11	6 10	eqtrd	⊢ ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 gcd ( 𝑀 · 𝑁 ) ) = 𝑀 )