Metamath Proof Explorer

Theorem gcd32

Description: Swap the second and third arguments of a gcd. (Contributed by Scott Fenton, 8-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion gcd32 
|- ( ( A e. ZZ /\ B e. ZZ /\ C e. ZZ ) -> ( ( A gcd B ) gcd C ) = ( ( A gcd C ) gcd B ) )

Proof

Step Hyp Ref Expression
1 gcdcom 
 |-  ( ( B e. ZZ /\ C e. ZZ ) -> ( B gcd C ) = ( C gcd B ) )
2 1 3adant1 
 |-  ( ( A e. ZZ /\ B e. ZZ /\ C e. ZZ ) -> ( B gcd C ) = ( C gcd B ) )
3 2 oveq2d 
 |-  ( ( A e. ZZ /\ B e. ZZ /\ C e. ZZ ) -> ( A gcd ( B gcd C ) ) = ( A gcd ( C gcd B ) ) )
4 gcdass 
 |-  ( ( A e. ZZ /\ B e. ZZ /\ C e. ZZ ) -> ( ( A gcd B ) gcd C ) = ( A gcd ( B gcd C ) ) )
5 gcdass 
 |-  ( ( A e. ZZ /\ C e. ZZ /\ B e. ZZ ) -> ( ( A gcd C ) gcd B ) = ( A gcd ( C gcd B ) ) )
6 5 3com23 
 |-  ( ( A e. ZZ /\ B e. ZZ /\ C e. ZZ ) -> ( ( A gcd C ) gcd B ) = ( A gcd ( C gcd B ) ) )
7 3 4 6 3eqtr4d 
 |-  ( ( A e. ZZ /\ B e. ZZ /\ C e. ZZ ) -> ( ( A gcd B ) gcd C ) = ( ( A gcd C ) gcd B ) )