Metamath Proof Explorer
Description: The gcd operator is commutative, deduction version. (Contributed by SN, 24-Aug-2024)
|
|
Ref |
Expression |
|
Hypotheses |
gcdcomd.m |
⊢ ( 𝜑 → 𝑀 ∈ ℤ ) |
|
|
gcdcomd.n |
⊢ ( 𝜑 → 𝑁 ∈ ℤ ) |
|
Assertion |
gcdcomd |
⊢ ( 𝜑 → ( 𝑀 gcd 𝑁 ) = ( 𝑁 gcd 𝑀 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
gcdcomd.m |
⊢ ( 𝜑 → 𝑀 ∈ ℤ ) |
2 |
|
gcdcomd.n |
⊢ ( 𝜑 → 𝑁 ∈ ℤ ) |
3 |
|
gcdcom |
⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 gcd 𝑁 ) = ( 𝑁 gcd 𝑀 ) ) |
4 |
1 2 3
|
syl2anc |
⊢ ( 𝜑 → ( 𝑀 gcd 𝑁 ) = ( 𝑁 gcd 𝑀 ) ) |