Metamath Proof Explorer


Theorem gcdnncl

Description: Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020)

Ref Expression
Assertion gcdnncl ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → ( 𝑀 gcd 𝑁 ) ∈ ℕ )

Proof

Step Hyp Ref Expression
1 simpl ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → 𝑀 ∈ ℕ )
2 1 nnzd ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → 𝑀 ∈ ℤ )
3 simpr ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → 𝑁 ∈ ℕ )
4 3 nnzd ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → 𝑁 ∈ ℤ )
5 3 nnne0d ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → 𝑁 ≠ 0 )
6 5 neneqd ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → ¬ 𝑁 = 0 )
7 6 intnand ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → ¬ ( 𝑀 = 0 ∧ 𝑁 = 0 ) )
8 gcdn0cl ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ¬ ( 𝑀 = 0 ∧ 𝑁 = 0 ) ) → ( 𝑀 gcd 𝑁 ) ∈ ℕ )
9 2 4 7 8 syl21anc ( ( 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → ( 𝑀 gcd 𝑁 ) ∈ ℕ )