Metamath Proof Explorer

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 ∈ ℕ 0
8 7 adantr ⊢ M ∈ ℕ ∧ N ∈ ℤ → M ∈ ℕ 0
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