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 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 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 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } = { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑁 ∧ 𝑛 ∥ 𝑀 ) }
4	3	supeq1i	⊢ sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } , ℝ , < ) = sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑁 ∧ 𝑛 ∥ 𝑀 ) } , ℝ , < )
5	1 4	ifbieq2i	⊢ if ( ( 𝑀 = 0 ∧ 𝑁 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } , ℝ , < ) ) = if ( ( 𝑁 = 0 ∧ 𝑀 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑁 ∧ 𝑛 ∥ 𝑀 ) } , ℝ , < ) )
6		gcdval	⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 gcd 𝑁 ) = if ( ( 𝑀 = 0 ∧ 𝑁 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } , ℝ , < ) ) )
7		gcdval	⊢ ( ( 𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ ) → ( 𝑁 gcd 𝑀 ) = if ( ( 𝑁 = 0 ∧ 𝑀 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑁 ∧ 𝑛 ∥ 𝑀 ) } , ℝ , < ) ) )
8	7	ancoms	⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑁 gcd 𝑀 ) = if ( ( 𝑁 = 0 ∧ 𝑀 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑁 ∧ 𝑛 ∥ 𝑀 ) } , ℝ , < ) ) )
9	5 6 8	3eqtr4a	⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 gcd 𝑁 ) = ( 𝑁 gcd 𝑀 ) )