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 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ ) → ( ( 𝐴 gcd 𝐵 ) gcd 𝐶 ) = ( ( 𝐴 gcd 𝐶 ) gcd 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		gcdcom	⊢ ( ( 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ ) → ( 𝐵 gcd 𝐶 ) = ( 𝐶 gcd 𝐵 ) )
2	1	3adant1	⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ ) → ( 𝐵 gcd 𝐶 ) = ( 𝐶 gcd 𝐵 ) )
3	2	oveq2d	⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ ) → ( 𝐴 gcd ( 𝐵 gcd 𝐶 ) ) = ( 𝐴 gcd ( 𝐶 gcd 𝐵 ) ) )
4		gcdass	⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ ) → ( ( 𝐴 gcd 𝐵 ) gcd 𝐶 ) = ( 𝐴 gcd ( 𝐵 gcd 𝐶 ) ) )
5		gcdass	⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐶 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → ( ( 𝐴 gcd 𝐶 ) gcd 𝐵 ) = ( 𝐴 gcd ( 𝐶 gcd 𝐵 ) ) )
6	5	3com23	⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ ) → ( ( 𝐴 gcd 𝐶 ) gcd 𝐵 ) = ( 𝐴 gcd ( 𝐶 gcd 𝐵 ) ) )
7	3 4 6	3eqtr4d	⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ ) → ( ( 𝐴 gcd 𝐵 ) gcd 𝐶 ) = ( ( 𝐴 gcd 𝐶 ) gcd 𝐵 ) )