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 → ( 𝑀 ‘ 𝑁 ) : ℙ ⟶ ℕ0 ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | 1arith.1 | ⊢ 𝑀 = ( 𝑛 ∈ ℕ ↦ ( 𝑝 ∈ ℙ ↦ ( 𝑝 pCnt 𝑛 ) ) ) | |
2 | 1 | 1arithlem1 | ⊢ ( 𝑁 ∈ ℕ → ( 𝑀 ‘ 𝑁 ) = ( 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 → ( 𝑀 ‘ 𝑁 ) : ℙ ⟶ ℕ0 ) |