| Step | Hyp | Ref
 | Expression | 
| 1 |   | simpl 109 | 
. . 3
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → 𝑃 ∈
ℙ) | 
| 2 |   | simprl 529 | 
. . 3
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → 𝑁 ∈
ℚ) | 
| 3 |   | ifnefalse 3572 | 
. . . . 5
⊢ (𝑁 ≠ 0 → if(𝑁 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) = (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) | 
| 4 | 3 | ad2antll 491 | 
. . . 4
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → if(𝑁 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) = (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) | 
| 5 |   | pcval.1 | 
. . . . . 6
⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < ) | 
| 6 |   | pcval.2 | 
. . . . . 6
⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < ) | 
| 7 | 5, 6 | pceu 12464 | 
. . . . 5
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → ∃!𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇))) | 
| 8 |   | euiotaex 5235 | 
. . . . 5
⊢
(∃!𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)) → (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇))) ∈ V) | 
| 9 | 7, 8 | syl 14 | 
. . . 4
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇))) ∈ V) | 
| 10 | 4, 9 | eqeltrd 2273 | 
. . 3
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → if(𝑁 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) ∈ V) | 
| 11 |   | simpr 110 | 
. . . . . 6
⊢ ((𝑝 = 𝑃 ∧ 𝑟 = 𝑁) → 𝑟 = 𝑁) | 
| 12 | 11 | eqeq1d 2205 | 
. . . . 5
⊢ ((𝑝 = 𝑃 ∧ 𝑟 = 𝑁) → (𝑟 = 0 ↔ 𝑁 = 0)) | 
| 13 |   | eqeq1 2203 | 
. . . . . . . 8
⊢ (𝑟 = 𝑁 → (𝑟 = (𝑥 / 𝑦) ↔ 𝑁 = (𝑥 / 𝑦))) | 
| 14 |   | oveq1 5929 | 
. . . . . . . . . . . . . 14
⊢ (𝑝 = 𝑃 → (𝑝↑𝑛) = (𝑃↑𝑛)) | 
| 15 | 14 | breq1d 4043 | 
. . . . . . . . . . . . 13
⊢ (𝑝 = 𝑃 → ((𝑝↑𝑛) ∥ 𝑥 ↔ (𝑃↑𝑛) ∥ 𝑥)) | 
| 16 | 15 | rabbidv 2752 | 
. . . . . . . . . . . 12
⊢ (𝑝 = 𝑃 → {𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥} = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}) | 
| 17 | 16 | supeq1d 7053 | 
. . . . . . . . . . 11
⊢ (𝑝 = 𝑃 → sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < )) | 
| 18 | 17, 5 | eqtr4di 2247 | 
. . . . . . . . . 10
⊢ (𝑝 = 𝑃 → sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) = 𝑆) | 
| 19 | 14 | breq1d 4043 | 
. . . . . . . . . . . . 13
⊢ (𝑝 = 𝑃 → ((𝑝↑𝑛) ∥ 𝑦 ↔ (𝑃↑𝑛) ∥ 𝑦)) | 
| 20 | 19 | rabbidv 2752 | 
. . . . . . . . . . . 12
⊢ (𝑝 = 𝑃 → {𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑦} = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}) | 
| 21 | 20 | supeq1d 7053 | 
. . . . . . . . . . 11
⊢ (𝑝 = 𝑃 → sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < ) = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < )) | 
| 22 | 21, 6 | eqtr4di 2247 | 
. . . . . . . . . 10
⊢ (𝑝 = 𝑃 → sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < ) = 𝑇) | 
| 23 | 18, 22 | oveq12d 5940 | 
. . . . . . . . 9
⊢ (𝑝 = 𝑃 → (sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0
∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < )) = (𝑆 − 𝑇)) | 
| 24 | 23 | eqeq2d 2208 | 
. . . . . . . 8
⊢ (𝑝 = 𝑃 → (𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0
∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < )) ↔ 𝑧 = (𝑆 − 𝑇))) | 
| 25 | 13, 24 | bi2anan9r 607 | 
. . . . . . 7
⊢ ((𝑝 = 𝑃 ∧ 𝑟 = 𝑁) → ((𝑟 = (𝑥 / 𝑦) ∧ 𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0
∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < ))) ↔ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) | 
| 26 | 25 | 2rexbidv 2522 | 
. . . . . 6
⊢ ((𝑝 = 𝑃 ∧ 𝑟 = 𝑁) → (∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑟 = (𝑥 / 𝑦) ∧ 𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0
∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < ))) ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) | 
| 27 | 26 | iotabidv 5241 | 
. . . . 5
⊢ ((𝑝 = 𝑃 ∧ 𝑟 = 𝑁) → (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑟 = (𝑥 / 𝑦) ∧ 𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0
∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < )))) = (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) | 
| 28 | 12, 27 | ifbieq2d 3585 | 
. . . 4
⊢ ((𝑝 = 𝑃 ∧ 𝑟 = 𝑁) → if(𝑟 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑟 = (𝑥 / 𝑦) ∧ 𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0
∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < ))))) = if(𝑁 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇))))) | 
| 29 |   | df-pc 12454 | 
. . . 4
⊢  pCnt =
(𝑝 ∈ ℙ, 𝑟 ∈ ℚ ↦ if(𝑟 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑟 = (𝑥 / 𝑦) ∧ 𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0
∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < )))))) | 
| 30 | 28, 29 | ovmpoga 6052 | 
. . 3
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℚ ∧ if(𝑁 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) ∈ V) → (𝑃 pCnt 𝑁) = if(𝑁 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇))))) | 
| 31 | 1, 2, 10, 30 | syl3anc 1249 | 
. 2
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = if(𝑁 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇))))) | 
| 32 | 31, 4 | eqtrd 2229 | 
1
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) |