Metamath Proof Explorer

Theorem gcdabs1

Description: gcd of the absolute value of the first operator. (Contributed by Scott Fenton, 2-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion gcdabs1 ⊢ N ∈ ℤ ∧ M ∈ ℤ → N gcd M = N gcd M

Proof

Step Hyp Ref Expression
1 oveq1 ⊢ N = N → N gcd M = N gcd M
2 1 a1i ⊢ N ∈ ℤ ∧ M ∈ ℤ → N = N → N gcd M = N gcd M
3 neggcd ⊢ N ∈ ℤ ∧ M ∈ ℤ → -N gcd M = N gcd M
4 oveq1 ⊢ N = − N → N gcd M = -N gcd M
5 4 eqeq1d ⊢ N = − N → N gcd M = N gcd M ↔ -N gcd M = N gcd M
6 3 5 syl5ibrcom ⊢ N ∈ ℤ ∧ M ∈ ℤ → N = − N → N gcd M = N gcd M
7 zre ⊢ N ∈ ℤ → N ∈ ℝ
8 7 absord ⊢ N ∈ ℤ → N = N ∨ N = − N
9 8 adantr ⊢ N ∈ ℤ ∧ M ∈ ℤ → N = N ∨ N = − N
10 2 6 9 mpjaod ⊢ N ∈ ℤ ∧ M ∈ ℤ → N gcd M = N gcd M