Metamath Proof Explorer
Description: A is equal to its gcd with B if and only if A divides B .
(Contributed by Mario Carneiro, 23-Feb-2014) (Proof shortened by AV, 8-Aug-2021)
|
|
Ref |
Expression |
|
Assertion |
gcdeq |
⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ) → ( ( 𝐴 gcd 𝐵 ) = 𝐴 ↔ 𝐴 ∥ 𝐵 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
nnz |
⊢ ( 𝐵 ∈ ℕ → 𝐵 ∈ ℤ ) |
2 |
|
gcdzeq |
⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ ) → ( ( 𝐴 gcd 𝐵 ) = 𝐴 ↔ 𝐴 ∥ 𝐵 ) ) |
3 |
1 2
|
sylan2 |
⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ) → ( ( 𝐴 gcd 𝐵 ) = 𝐴 ↔ 𝐴 ∥ 𝐵 ) ) |