Theorem gcdcom

Description: The gcd operator is commutative. Theorem 1.4(a) in ApostolNT p. 16. (Contributed by Paul Chapman, 21-Mar-2011)

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

Proof

Step	Hyp	Ref	Expression
1		ancom	|- ( ( M = 0 /\ N = 0 ) <-> ( N = 0 /\ M = 0 ) )
2		ancom	|- ( ( n || M /\ n || N ) <-> ( n || N /\ n || M ) )
3	2	rabbii	|- { n e. ZZ | ( n || M /\ n || N ) } = { n e. ZZ | ( n || N /\ n || M ) }
4	3	supeq1i	|- sup ( { n e. ZZ | ( n || M /\ n || N ) } , RR , < ) = sup ( { n e. ZZ | ( n || N /\ n || M ) } , RR , < )
5	1 4	ifbieq2i	|- if ( ( M = 0 /\ N = 0 ) , 0 , sup ( { n e. ZZ | ( n || M /\ n || N ) } , RR , < ) ) = if ( ( N = 0 /\ M = 0 ) , 0 , sup ( { n e. ZZ | ( n || N /\ n || M ) } , RR , < ) )
6		gcdval	|- ( ( M e. ZZ /\ N e. ZZ ) -> ( M gcd N ) = if ( ( M = 0 /\ N = 0 ) , 0 , sup ( { n e. ZZ | ( n || M /\ n || N ) } , RR , < ) ) )
7		gcdval	|- ( ( N e. ZZ /\ M e. ZZ ) -> ( N gcd M ) = if ( ( N = 0 /\ M = 0 ) , 0 , sup ( { n e. ZZ | ( n || N /\ n || M ) } , RR , < ) ) )
8	7	ancoms	|- ( ( M e. ZZ /\ N e. ZZ ) -> ( N gcd M ) = if ( ( N = 0 /\ M = 0 ) , 0 , sup ( { n e. ZZ | ( n || N /\ n || M ) } , RR , < ) ) )
9	5 6 8	3eqtr4a	|- ( ( M e. ZZ /\ N e. ZZ ) -> ( M gcd N ) = ( N gcd M ) )