Metamath Proof Explorer


Theorem 1arithlem3

Description: Lemma for 1arith . (Contributed by Mario Carneiro, 30-May-2014)

Ref Expression
Hypothesis 1arith.1 𝑀 = ( 𝑛 ∈ ℕ ↦ ( 𝑝 ∈ ℙ ↦ ( 𝑝 pCnt 𝑛 ) ) )
Assertion 1arithlem3 ( 𝑁 ∈ ℕ → ( 𝑀𝑁 ) : ℙ ⟶ ℕ0 )

Proof

Step Hyp Ref Expression
1 1arith.1 𝑀 = ( 𝑛 ∈ ℕ ↦ ( 𝑝 ∈ ℙ ↦ ( 𝑝 pCnt 𝑛 ) ) )
2 1 1arithlem1 ( 𝑁 ∈ ℕ → ( 𝑀𝑁 ) = ( 𝑝 ∈ ℙ ↦ ( 𝑝 pCnt 𝑁 ) ) )
3 pccl ( ( 𝑝 ∈ ℙ ∧ 𝑁 ∈ ℕ ) → ( 𝑝 pCnt 𝑁 ) ∈ ℕ0 )
4 3 ancoms ( ( 𝑁 ∈ ℕ ∧ 𝑝 ∈ ℙ ) → ( 𝑝 pCnt 𝑁 ) ∈ ℕ0 )
5 2 4 fmpt3d ( 𝑁 ∈ ℕ → ( 𝑀𝑁 ) : ℙ ⟶ ℕ0 )