Proof of Theorem divalg
| Step | Hyp | Ref
| Expression |
| 1 | | eqeq1 2739 |
. . . . . 6
⊢ (𝑁 = if(𝑁 ∈ ℤ, 𝑁, 1) → (𝑁 = ((𝑞 · 𝐷) + 𝑟) ↔ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · 𝐷) + 𝑟))) |
| 2 | 1 | 3anbi3d 1444 |
. . . . 5
⊢ (𝑁 = if(𝑁 ∈ ℤ, 𝑁, 1) → ((0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)) ↔ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · 𝐷) + 𝑟)))) |
| 3 | 2 | rexbidv 3164 |
. . . 4
⊢ (𝑁 = if(𝑁 ∈ ℤ, 𝑁, 1) → (∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)) ↔ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · 𝐷) + 𝑟)))) |
| 4 | 3 | reubidv 3377 |
. . 3
⊢ (𝑁 = if(𝑁 ∈ ℤ, 𝑁, 1) → (∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)) ↔ ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · 𝐷) + 𝑟)))) |
| 5 | | fveq2 6875 |
. . . . . . 7
⊢ (𝐷 = if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) → (abs‘𝐷) = (abs‘if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1))) |
| 6 | 5 | breq2d 5131 |
. . . . . 6
⊢ (𝐷 = if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) → (𝑟 < (abs‘𝐷) ↔ 𝑟 < (abs‘if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)))) |
| 7 | | oveq2 7411 |
. . . . . . . 8
⊢ (𝐷 = if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) → (𝑞 · 𝐷) = (𝑞 · if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1))) |
| 8 | 7 | oveq1d 7418 |
. . . . . . 7
⊢ (𝐷 = if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) → ((𝑞 · 𝐷) + 𝑟) = ((𝑞 · if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)) + 𝑟)) |
| 9 | 8 | eqeq2d 2746 |
. . . . . 6
⊢ (𝐷 = if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) → (if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · 𝐷) + 𝑟) ↔ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)) + 𝑟))) |
| 10 | 6, 9 | 3anbi23d 1441 |
. . . . 5
⊢ (𝐷 = if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) → ((0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · 𝐷) + 𝑟)) ↔ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)) ∧ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)) + 𝑟)))) |
| 11 | 10 | rexbidv 3164 |
. . . 4
⊢ (𝐷 = if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) → (∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · 𝐷) + 𝑟)) ↔ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)) ∧ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)) + 𝑟)))) |
| 12 | 11 | reubidv 3377 |
. . 3
⊢ (𝐷 = if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) → (∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · 𝐷) + 𝑟)) ↔ ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)) ∧ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)) + 𝑟)))) |
| 13 | | 1z 12620 |
. . . . 5
⊢ 1 ∈
ℤ |
| 14 | 13 | elimel 4570 |
. . . 4
⊢ if(𝑁 ∈ ℤ, 𝑁, 1) ∈
ℤ |
| 15 | | simpl 482 |
. . . . 5
⊢ ((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → 𝐷 ∈
ℤ) |
| 16 | | eleq1 2822 |
. . . . 5
⊢ (𝐷 = if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) → (𝐷 ∈ ℤ ↔ if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) ∈ ℤ)) |
| 17 | | eleq1 2822 |
. . . . 5
⊢ (1 =
if((𝐷 ∈ ℤ ∧
𝐷 ≠ 0), 𝐷, 1) → (1 ∈ ℤ
↔ if((𝐷 ∈ ℤ
∧ 𝐷 ≠ 0), 𝐷, 1) ∈
ℤ)) |
| 18 | 15, 16, 17, 13 | elimdhyp 4571 |
. . . 4
⊢ if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) ∈ ℤ |
| 19 | | simpr 484 |
. . . . 5
⊢ ((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → 𝐷 ≠ 0) |
| 20 | | neeq1 2994 |
. . . . 5
⊢ (𝐷 = if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) → (𝐷 ≠ 0 ↔ if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) ≠ 0)) |
| 21 | | neeq1 2994 |
. . . . 5
⊢ (1 =
if((𝐷 ∈ ℤ ∧
𝐷 ≠ 0), 𝐷, 1) → (1 ≠ 0 ↔
if((𝐷 ∈ ℤ ∧
𝐷 ≠ 0), 𝐷, 1) ≠ 0)) |
| 22 | | ax-1ne0 11196 |
. . . . 5
⊢ 1 ≠
0 |
| 23 | 19, 20, 21, 22 | elimdhyp 4571 |
. . . 4
⊢ if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) ≠ 0 |
| 24 | | eqid 2735 |
. . . 4
⊢ {𝑟 ∈ ℕ0
∣ if((𝐷 ∈
ℤ ∧ 𝐷 ≠ 0),
𝐷, 1) ∥ (if(𝑁 ∈ ℤ, 𝑁, 1) − 𝑟)} = {𝑟 ∈ ℕ0 ∣ if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1) ∥ (if(𝑁 ∈ ℤ, 𝑁, 1) − 𝑟)} |
| 25 | 14, 18, 23, 24 | divalglem10 16419 |
. . 3
⊢
∃!𝑟 ∈
ℤ ∃𝑞 ∈
ℤ (0 ≤ 𝑟 ∧
𝑟 < (abs‘if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)) ∧ if(𝑁 ∈ ℤ, 𝑁, 1) = ((𝑞 · if((𝐷 ∈ ℤ ∧ 𝐷 ≠ 0), 𝐷, 1)) + 𝑟)) |
| 26 | 4, 12, 25 | dedth2h 4560 |
. 2
⊢ ((𝑁 ∈ ℤ ∧ (𝐷 ∈ ℤ ∧ 𝐷 ≠ 0)) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟))) |
| 27 | 26 | 3impb 1114 |
1
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟))) |