Theorem gcdf

Description: Domain and codomain of the gcd operator. (Contributed by Paul Chapman, 31-Mar-2011) (Revised by Mario Carneiro, 16-Nov-2013)

	Ref	Expression
Assertion	gcdf	⊢ gcd : ( ℤ × ℤ ) ⟶ ℕ0

Proof

Step	Hyp	Ref	Expression
1		gcdval	⊢ ( ( 𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ ) → ( 𝑥 gcd 𝑦 ) = if ( ( 𝑥 = 0 ∧ 𝑦 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) ) )
2		gcdcl	⊢ ( ( 𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ ) → ( 𝑥 gcd 𝑦 ) ∈ ℕ0 )
3	1 2	eqeltrrd	⊢ ( ( 𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ ) → if ( ( 𝑥 = 0 ∧ 𝑦 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) ) ∈ ℕ0 )
4	3	rgen2	⊢ ∀ 𝑥 ∈ ℤ ∀ 𝑦 ∈ ℤ if ( ( 𝑥 = 0 ∧ 𝑦 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) ) ∈ ℕ0
5		df-gcd	⊢ gcd = ( 𝑥 ∈ ℤ , 𝑦 ∈ ℤ ↦ if ( ( 𝑥 = 0 ∧ 𝑦 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) ) )
6	5	fmpo	⊢ ( ∀ 𝑥 ∈ ℤ ∀ 𝑦 ∈ ℤ if ( ( 𝑥 = 0 ∧ 𝑦 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) ) ∈ ℕ0 ↔ gcd : ( ℤ × ℤ ) ⟶ ℕ0 )
7	4 6	mpbi	⊢ gcd : ( ℤ × ℤ ) ⟶ ℕ0