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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 ) → ( ( abs ‘ 𝑁 ) gcd 𝑀 ) = ( 𝑁 gcd 𝑀 ) )

Proof

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