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Metamath Proof Explorer

Theorem gcdabs2

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

Ref Expression
Assertion gcdabs2 
|- ( ( N e. ZZ /\ M e. ZZ ) -> ( N gcd ( abs ` M ) ) = ( N gcd M ) )

Proof

Step Hyp Ref Expression
1 gcdabs1 
 |-  ( ( M e. ZZ /\ N e. ZZ ) -> ( ( abs ` M ) gcd N ) = ( M gcd N ) )
2 1 ancoms 
 |-  ( ( N e. ZZ /\ M e. ZZ ) -> ( ( abs ` M ) gcd N ) = ( M gcd N ) )
3 zabscl 
 |-  ( M e. ZZ -> ( abs ` M ) e. ZZ )
4 gcdcom 
 |-  ( ( N e. ZZ /\ ( abs ` M ) e. ZZ ) -> ( N gcd ( abs ` M ) ) = ( ( abs ` M ) gcd N ) )
5 3 4 sylan2 
 |-  ( ( N e. ZZ /\ M e. ZZ ) -> ( N gcd ( abs ` M ) ) = ( ( abs ` M ) gcd N ) )
6 gcdcom 
 |-  ( ( N e. ZZ /\ M e. ZZ ) -> ( N gcd M ) = ( M gcd N ) )
7 2 5 6 3eqtr4d 
 |-  ( ( N e. ZZ /\ M e. ZZ ) -> ( N gcd ( abs ` M ) ) = ( N gcd M ) )