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Metamath Proof Explorer

Theorem nndvdslegcd

Description: A positive integer which divides both positive operands of the gcd operator is bounded by it. (Contributed by AV, 9-Aug-2020)

Ref Expression
Assertion nndvdslegcd ( ( 𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → ( ( 𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁 ) → 𝐾 ≤ ( 𝑀 gcd 𝑁 ) ) )

Proof

Step Hyp Ref Expression
1 nnz ⊢ ( 𝐾 ∈ ℕ → 𝐾 ∈ ℤ )
2 nnz ⊢ ( 𝑀 ∈ ℕ → 𝑀 ∈ ℤ )
3 nnz ⊢ ( 𝑁 ∈ ℕ → 𝑁 ∈ ℤ )
4 1 2 3 3anim123i ⊢ ( ( 𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → ( 𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) )
5 nnne0 ⊢ ( 𝑀 ∈ ℕ → 𝑀 ≠ 0 )
6 5 neneqd ⊢ ( 𝑀 ∈ ℕ → ¬ 𝑀 = 0 )
7 6 3ad2ant2 ⊢ ( ( 𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → ¬ 𝑀 = 0 )
8 7 intnanrd ⊢ ( ( 𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → ¬ ( 𝑀 = 0 ∧ 𝑁 = 0 ) )
9 dvdslegcd ⊢ ( ( ( 𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ¬ ( 𝑀 = 0 ∧ 𝑁 = 0 ) ) → ( ( 𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁 ) → 𝐾 ≤ ( 𝑀 gcd 𝑁 ) ) )
10 4 8 9 syl2anc ⊢ ( ( 𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ) → ( ( 𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁 ) → 𝐾 ≤ ( 𝑀 gcd 𝑁 ) ) )