Metamath Proof Explorer


Theorem isppw

Description: Two ways to say that A is a prime power. (Contributed by Mario Carneiro, 7-Apr-2016)

Ref Expression
Assertion isppw ⊢ A ∈ ℕ → Λ ⁡ A ≠ 0 ↔ ∃! p ∈ ℙ p ∥ A

Proof

Step Hyp Ref Expression
1 eqid ⊢ p ∈ ℙ | p ∥ A = p ∈ ℙ | p ∥ A
2 1 vmaval ⊢ A ∈ ℕ → Λ ⁡ A = if p ∈ ℙ | p ∥ A = 1 log ⁡ ⋃ p ∈ ℙ | p ∥ A 0
3 2 neeq1d ⊢ A ∈ ℕ → Λ ⁡ A ≠ 0 ↔ if p ∈ ℙ | p ∥ A = 1 log ⁡ ⋃ p ∈ ℙ | p ∥ A 0 ≠ 0
4 reuen1 ⊢ ∃! p ∈ ℙ p ∥ A ↔ p ∈ ℙ | p ∥ A ≈ 1 𝑜
5 hash1 ⊢ 1 𝑜 = 1
6 5 eqeq2i ⊢ p ∈ ℙ | p ∥ A = 1 𝑜 ↔ p ∈ ℙ | p ∥ A = 1
7 prmdvdsfi ⊢ A ∈ ℕ → p ∈ ℙ | p ∥ A ∈ Fin
8 1onn ⊢ 1 𝑜 ∈ ω
9 nnfi ⊢ 1 𝑜 ∈ ω → 1 𝑜 ∈ Fin
10 8 9 ax-mp ⊢ 1 𝑜 ∈ Fin
11 hashen ⊢ p ∈ ℙ | p ∥ A ∈ Fin ∧ 1 𝑜 ∈ Fin → p ∈ ℙ | p ∥ A = 1 𝑜 ↔ p ∈ ℙ | p ∥ A ≈ 1 𝑜
12 7 10 11 sylancl ⊢ A ∈ ℕ → p ∈ ℙ | p ∥ A = 1 𝑜 ↔ p ∈ ℙ | p ∥ A ≈ 1 𝑜
13 6 12 bitr3id ⊢ A ∈ ℕ → p ∈ ℙ | p ∥ A = 1 ↔ p ∈ ℙ | p ∥ A ≈ 1 𝑜
14 13 biimpar ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → p ∈ ℙ | p ∥ A = 1
15 14 iftrued ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → if p ∈ ℙ | p ∥ A = 1 log ⁡ ⋃ p ∈ ℙ | p ∥ A 0 = log ⁡ ⋃ p ∈ ℙ | p ∥ A
16 en1b ⊢ p ∈ ℙ | p ∥ A ≈ 1 𝑜 ↔ p ∈ ℙ | p ∥ A = ⋃ p ∈ ℙ | p ∥ A
17 16 bilani ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → p ∈ ℙ | p ∥ A = ⋃ p ∈ ℙ | p ∥ A
18 ssrab2 ⊢ p ∈ ℙ | p ∥ A ⊆ ℙ
19 17 18 eqsstrrdi ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → ⋃ p ∈ ℙ | p ∥ A ⊆ ℙ
20 7 uniexd ⊢ A ∈ ℕ → ⋃ p ∈ ℙ | p ∥ A ∈ V
21 20 adantr ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → ⋃ p ∈ ℙ | p ∥ A ∈ V
22 snssg ⊢ ⋃ p ∈ ℙ | p ∥ A ∈ V → ⋃ p ∈ ℙ | p ∥ A ∈ ℙ ↔ ⋃ p ∈ ℙ | p ∥ A ⊆ ℙ
23 21 22 syl ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → ⋃ p ∈ ℙ | p ∥ A ∈ ℙ ↔ ⋃ p ∈ ℙ | p ∥ A ⊆ ℙ
24 19 23 mpbird ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → ⋃ p ∈ ℙ | p ∥ A ∈ ℙ
25 prmuz2 ⊢ ⋃ p ∈ ℙ | p ∥ A ∈ ℙ → ⋃ p ∈ ℙ | p ∥ A ∈ ℤ ≥ 2
26 24 25 syl ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → ⋃ p ∈ ℙ | p ∥ A ∈ ℤ ≥ 2
27 eluzelre ⊢ ⋃ p ∈ ℙ | p ∥ A ∈ ℤ ≥ 2 → ⋃ p ∈ ℙ | p ∥ A ∈ ℝ
28 26 27 syl ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → ⋃ p ∈ ℙ | p ∥ A ∈ ℝ
29 eluz2gt1 ⊢ ⋃ p ∈ ℙ | p ∥ A ∈ ℤ ≥ 2 → 1 < ⋃ p ∈ ℙ | p ∥ A
30 26 29 syl ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → 1 < ⋃ p ∈ ℙ | p ∥ A
31 28 30 rplogcld ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → log ⁡ ⋃ p ∈ ℙ | p ∥ A ∈ ℝ +
32 31 rpne0d ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → log ⁡ ⋃ p ∈ ℙ | p ∥ A ≠ 0
33 15 32 eqnetrd ⊢ A ∈ ℕ ∧ p ∈ ℙ | p ∥ A ≈ 1 𝑜 → if p ∈ ℙ | p ∥ A = 1 log ⁡ ⋃ p ∈ ℙ | p ∥ A 0 ≠ 0
34 33 ex ⊢ A ∈ ℕ → p ∈ ℙ | p ∥ A ≈ 1 𝑜 → if p ∈ ℙ | p ∥ A = 1 log ⁡ ⋃ p ∈ ℙ | p ∥ A 0 ≠ 0
35 iffalse ⊢ ¬ p ∈ ℙ | p ∥ A = 1 → if p ∈ ℙ | p ∥ A = 1 log ⁡ ⋃ p ∈ ℙ | p ∥ A 0 = 0
36 35 necon1ai ⊢ if p ∈ ℙ | p ∥ A = 1 log ⁡ ⋃ p ∈ ℙ | p ∥ A 0 ≠ 0 → p ∈ ℙ | p ∥ A = 1
37 36 13 imbitrid ⊢ A ∈ ℕ → if p ∈ ℙ | p ∥ A = 1 log ⁡ ⋃ p ∈ ℙ | p ∥ A 0 ≠ 0 → p ∈ ℙ | p ∥ A ≈ 1 𝑜
38 34 37 impbid ⊢ A ∈ ℕ → p ∈ ℙ | p ∥ A ≈ 1 𝑜 ↔ if p ∈ ℙ | p ∥ A = 1 log ⁡ ⋃ p ∈ ℙ | p ∥ A 0 ≠ 0
39 4 38 bitrid ⊢ A ∈ ℕ → ∃! p ∈ ℙ p ∥ A ↔ if p ∈ ℙ | p ∥ A = 1 log ⁡ ⋃ p ∈ ℙ | p ∥ A 0 ≠ 0
40 3 39 bitr4d ⊢ A ∈ ℕ → Λ ⁡ A ≠ 0 ↔ ∃! p ∈ ℙ p ∥ A