Metamath Proof Explorer
Description: Adding a multiple of one operand of the gcd operator to the other does
not alter the result. (Contributed by metakunt, 25-Apr-2024)
|
|
Ref |
Expression |
|
Hypotheses |
gcdaddmzz2nni.1 |
⊢ 𝑀 ∈ ℕ |
|
|
gcdaddmzz2nni.2 |
⊢ 𝑁 ∈ ℕ |
|
|
gcdaddmzz2nni.3 |
⊢ 𝐾 ∈ ℤ |
|
Assertion |
gcdaddmzz2nni |
⊢ ( 𝑀 gcd 𝑁 ) = ( 𝑀 gcd ( 𝑁 + ( 𝐾 · 𝑀 ) ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
gcdaddmzz2nni.1 |
⊢ 𝑀 ∈ ℕ |
2 |
|
gcdaddmzz2nni.2 |
⊢ 𝑁 ∈ ℕ |
3 |
|
gcdaddmzz2nni.3 |
⊢ 𝐾 ∈ ℤ |
4 |
1
|
nnzi |
⊢ 𝑀 ∈ ℤ |
5 |
2
|
nnzi |
⊢ 𝑁 ∈ ℤ |
6 |
3 4 5
|
3pm3.2i |
⊢ ( 𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) |
7 |
|
gcdaddm |
⊢ ( ( 𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 gcd 𝑁 ) = ( 𝑀 gcd ( 𝑁 + ( 𝐾 · 𝑀 ) ) ) ) |
8 |
6 7
|
ax-mp |
⊢ ( 𝑀 gcd 𝑁 ) = ( 𝑀 gcd ( 𝑁 + ( 𝐾 · 𝑀 ) ) ) |