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 ) |