Theorem bezoutr

Description: Partial converse to bezout . Existence of a linear combination does not set the GCD, but it does upper bound it. (Contributed by Stefan O'Rear, 23-Sep-2014)

	Ref	Expression
Assertion	bezoutr	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → ( 𝐴 gcd 𝐵 ) ∥ ( ( 𝐴 · 𝑋 ) + ( 𝐵 · 𝑌 ) ) )

Proof

Step	Hyp	Ref	Expression
1		gcdcl	⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → ( 𝐴 gcd 𝐵 ) ∈ ℕ0 )
2	1	nn0zd	⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → ( 𝐴 gcd 𝐵 ) ∈ ℤ )
3	2	adantr	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → ( 𝐴 gcd 𝐵 ) ∈ ℤ )
4		simpll	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → 𝐴 ∈ ℤ )
5		simprl	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → 𝑋 ∈ ℤ )
6	4 5	zmulcld	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → ( 𝐴 · 𝑋 ) ∈ ℤ )
7		simplr	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → 𝐵 ∈ ℤ )
8		simprr	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → 𝑌 ∈ ℤ )
9	7 8	zmulcld	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → ( 𝐵 · 𝑌 ) ∈ ℤ )
10		gcddvds	⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → ( ( 𝐴 gcd 𝐵 ) ∥ 𝐴 ∧ ( 𝐴 gcd 𝐵 ) ∥ 𝐵 ) )
11	10	adantr	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → ( ( 𝐴 gcd 𝐵 ) ∥ 𝐴 ∧ ( 𝐴 gcd 𝐵 ) ∥ 𝐵 ) )
12	11	simpld	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → ( 𝐴 gcd 𝐵 ) ∥ 𝐴 )
13	3 4 5 12	dvdsmultr1d	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → ( 𝐴 gcd 𝐵 ) ∥ ( 𝐴 · 𝑋 ) )
14	11	simprd	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → ( 𝐴 gcd 𝐵 ) ∥ 𝐵 )
15	3 7 8 14	dvdsmultr1d	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → ( 𝐴 gcd 𝐵 ) ∥ ( 𝐵 · 𝑌 ) )
16	3 6 9 13 15	dvds2addd	⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ) ) → ( 𝐴 gcd 𝐵 ) ∥ ( ( 𝐴 · 𝑋 ) + ( 𝐵 · 𝑌 ) ) )