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Description: Closure of the divisor function. (Contributed by Mario Carneiro, 21-Jun-2015)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | sgmnncl | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → ( 𝐴 σ 𝐵 ) ∈ ℕ ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | nn0z | ⊢ ( 𝐴 ∈ ℕ0 → 𝐴 ∈ ℤ ) | |
| 2 | sgmval2 | ⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ ) → ( 𝐴 σ 𝐵 ) = Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ) | |
| 3 | 1 2 | sylan | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → ( 𝐴 σ 𝐵 ) = Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ) |
| 4 | fzfid | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → ( 1 ... 𝐵 ) ∈ Fin ) | |
| 5 | dvdsssfz1 | ⊢ ( 𝐵 ∈ ℕ → { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ⊆ ( 1 ... 𝐵 ) ) | |
| 6 | 5 | adantl | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ⊆ ( 1 ... 𝐵 ) ) |
| 7 | 4 6 | ssfid | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ∈ Fin ) |
| 8 | elrabi | ⊢ ( 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } → 𝑘 ∈ ℕ ) | |
| 9 | simpl | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → 𝐴 ∈ ℕ0 ) | |
| 10 | nnexpcl | ⊢ ( ( 𝑘 ∈ ℕ ∧ 𝐴 ∈ ℕ0 ) → ( 𝑘 ↑ 𝐴 ) ∈ ℕ ) | |
| 11 | 8 9 10 | syl2anr | ⊢ ( ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) ∧ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ) → ( 𝑘 ↑ 𝐴 ) ∈ ℕ ) |
| 12 | 11 | nnzd | ⊢ ( ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) ∧ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ) → ( 𝑘 ↑ 𝐴 ) ∈ ℤ ) |
| 13 | 7 12 | fsumzcl | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ∈ ℤ ) |
| 14 | nnz | ⊢ ( 𝐵 ∈ ℕ → 𝐵 ∈ ℤ ) | |
| 15 | iddvds | ⊢ ( 𝐵 ∈ ℤ → 𝐵 ∥ 𝐵 ) | |
| 16 | 14 15 | syl | ⊢ ( 𝐵 ∈ ℕ → 𝐵 ∥ 𝐵 ) |
| 17 | breq1 | ⊢ ( 𝑝 = 𝐵 → ( 𝑝 ∥ 𝐵 ↔ 𝐵 ∥ 𝐵 ) ) | |
| 18 | 17 | rspcev | ⊢ ( ( 𝐵 ∈ ℕ ∧ 𝐵 ∥ 𝐵 ) → ∃ 𝑝 ∈ ℕ 𝑝 ∥ 𝐵 ) |
| 19 | 16 18 | mpdan | ⊢ ( 𝐵 ∈ ℕ → ∃ 𝑝 ∈ ℕ 𝑝 ∥ 𝐵 ) |
| 20 | rabn0 | ⊢ ( { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ≠ ∅ ↔ ∃ 𝑝 ∈ ℕ 𝑝 ∥ 𝐵 ) | |
| 21 | 19 20 | sylibr | ⊢ ( 𝐵 ∈ ℕ → { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ≠ ∅ ) |
| 22 | 21 | adantl | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ≠ ∅ ) |
| 23 | 11 | nnrpd | ⊢ ( ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) ∧ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ) → ( 𝑘 ↑ 𝐴 ) ∈ ℝ+ ) |
| 24 | 7 22 23 | fsumrpcl | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ∈ ℝ+ ) |
| 25 | 24 | rpgt0d | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → 0 < Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ) |
| 26 | elnnz | ⊢ ( Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ∈ ℕ ↔ ( Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ∈ ℤ ∧ 0 < Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ) ) | |
| 27 | 13 25 26 | sylanbrc | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ∈ ℕ ) |
| 28 | 3 27 | eqeltrd | ⊢ ( ( 𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ ) → ( 𝐴 σ 𝐵 ) ∈ ℕ ) |