Theorem gcdval

Description: The value of the gcd operator. ( M gcd N ) is the greatest common divisor of M and N . If M and N are both 0 , the result is defined conventionally as 0 . (Contributed by Paul Chapman, 21-Mar-2011) (Revised by Mario Carneiro, 10-Nov-2013)

	Ref	Expression
Assertion	gcdval	⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 gcd 𝑁 ) = if ( ( 𝑀 = 0 ∧ 𝑁 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } , ℝ , < ) ) )

Proof

Step	Hyp	Ref	Expression
1		eqeq1	⊢ ( 𝑥 = 𝑀 → ( 𝑥 = 0 ↔ 𝑀 = 0 ) )
2	1	anbi1d	⊢ ( 𝑥 = 𝑀 → ( ( 𝑥 = 0 ∧ 𝑦 = 0 ) ↔ ( 𝑀 = 0 ∧ 𝑦 = 0 ) ) )
3		breq2	⊢ ( 𝑥 = 𝑀 → ( 𝑛 ∥ 𝑥 ↔ 𝑛 ∥ 𝑀 ) )
4	3	anbi1d	⊢ ( 𝑥 = 𝑀 → ( ( 𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦 ) ↔ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑦 ) ) )
5	4	rabbidv	⊢ ( 𝑥 = 𝑀 → { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦 ) } = { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑦 ) } )
6	5	supeq1d	⊢ ( 𝑥 = 𝑀 → sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) = sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) )
7	2 6	ifbieq2d	⊢ ( 𝑥 = 𝑀 → if ( ( 𝑥 = 0 ∧ 𝑦 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) ) = if ( ( 𝑀 = 0 ∧ 𝑦 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) ) )
8		eqeq1	⊢ ( 𝑦 = 𝑁 → ( 𝑦 = 0 ↔ 𝑁 = 0 ) )
9	8	anbi2d	⊢ ( 𝑦 = 𝑁 → ( ( 𝑀 = 0 ∧ 𝑦 = 0 ) ↔ ( 𝑀 = 0 ∧ 𝑁 = 0 ) ) )
10		breq2	⊢ ( 𝑦 = 𝑁 → ( 𝑛 ∥ 𝑦 ↔ 𝑛 ∥ 𝑁 ) )
11	10	anbi2d	⊢ ( 𝑦 = 𝑁 → ( ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑦 ) ↔ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) ) )
12	11	rabbidv	⊢ ( 𝑦 = 𝑁 → { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑦 ) } = { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } )
13	12	supeq1d	⊢ ( 𝑦 = 𝑁 → sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) = sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } , ℝ , < ) )
14	9 13	ifbieq2d	⊢ ( 𝑦 = 𝑁 → if ( ( 𝑀 = 0 ∧ 𝑦 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) ) = if ( ( 𝑀 = 0 ∧ 𝑁 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } , ℝ , < ) ) )
15		df-gcd	⊢ gcd = ( 𝑥 ∈ ℤ , 𝑦 ∈ ℤ ↦ if ( ( 𝑥 = 0 ∧ 𝑦 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦 ) } , ℝ , < ) ) )
16		c0ex	⊢ 0 ∈ V
17		ltso	⊢ < Or ℝ
18	17	supex	⊢ sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } , ℝ , < ) ∈ V
19	16 18	ifex	⊢ if ( ( 𝑀 = 0 ∧ 𝑁 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } , ℝ , < ) ) ∈ V
20	7 14 15 19	ovmpo	⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 gcd 𝑁 ) = if ( ( 𝑀 = 0 ∧ 𝑁 = 0 ) , 0 , sup ( { 𝑛 ∈ ℤ ∣ ( 𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁 ) } , ℝ , < ) ) )