Proof of Theorem pythagtriplem17
Step | Hyp | Ref
| Expression |
1 | | pythagtriplem15.1 |
. . . . 5
⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) |
2 | 1 | pythagtriplem12 15369 |
. . . 4
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑀↑2) = ((𝐶 + 𝐴) / 2)) |
3 | | pythagtriplem15.2 |
. . . . 5
⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) |
4 | 3 | pythagtriplem14 15371 |
. . . 4
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑁↑2) = ((𝐶 − 𝐴) / 2)) |
5 | 2, 4 | oveq12d 6567 |
. . 3
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ((𝑀↑2) + (𝑁↑2)) = (((𝐶 + 𝐴) / 2) + ((𝐶 − 𝐴) / 2))) |
6 | | nncn 10905 |
. . . . . . 7
⊢ (𝐶 ∈ ℕ → 𝐶 ∈
ℂ) |
7 | 6 | 3ad2ant3 1077 |
. . . . . 6
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) → 𝐶 ∈
ℂ) |
8 | 7 | 3ad2ant1 1075 |
. . . . 5
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐶 ∈ ℂ) |
9 | | nncn 10905 |
. . . . . . 7
⊢ (𝐴 ∈ ℕ → 𝐴 ∈
ℂ) |
10 | 9 | 3ad2ant1 1075 |
. . . . . 6
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) → 𝐴 ∈
ℂ) |
11 | 10 | 3ad2ant1 1075 |
. . . . 5
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐴 ∈ ℂ) |
12 | 8, 11 | addcld 9938 |
. . . 4
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝐶 + 𝐴) ∈ ℂ) |
13 | 8, 11 | subcld 10271 |
. . . 4
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝐶 − 𝐴) ∈ ℂ) |
14 | | 2cnne0 11119 |
. . . . 5
⊢ (2 ∈
ℂ ∧ 2 ≠ 0) |
15 | | divdir 10589 |
. . . . 5
⊢ (((𝐶 + 𝐴) ∈ ℂ ∧ (𝐶 − 𝐴) ∈ ℂ ∧ (2 ∈ ℂ
∧ 2 ≠ 0)) → (((𝐶 + 𝐴) + (𝐶 − 𝐴)) / 2) = (((𝐶 + 𝐴) / 2) + ((𝐶 − 𝐴) / 2))) |
16 | 14, 15 | mp3an3 1405 |
. . . 4
⊢ (((𝐶 + 𝐴) ∈ ℂ ∧ (𝐶 − 𝐴) ∈ ℂ) → (((𝐶 + 𝐴) + (𝐶 − 𝐴)) / 2) = (((𝐶 + 𝐴) / 2) + ((𝐶 − 𝐴) / 2))) |
17 | 12, 13, 16 | syl2anc 691 |
. . 3
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (((𝐶 + 𝐴) + (𝐶 − 𝐴)) / 2) = (((𝐶 + 𝐴) / 2) + ((𝐶 − 𝐴) / 2))) |
18 | 5, 17 | eqtr4d 2647 |
. 2
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ((𝑀↑2) + (𝑁↑2)) = (((𝐶 + 𝐴) + (𝐶 − 𝐴)) / 2)) |
19 | 8, 11, 8 | ppncand 10311 |
. . . 4
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ((𝐶 + 𝐴) + (𝐶 − 𝐴)) = (𝐶 + 𝐶)) |
20 | 8 | 2timesd 11152 |
. . . 4
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (2 · 𝐶) = (𝐶 + 𝐶)) |
21 | 19, 20 | eqtr4d 2647 |
. . 3
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ((𝐶 + 𝐴) + (𝐶 − 𝐴)) = (2 · 𝐶)) |
22 | 21 | oveq1d 6564 |
. 2
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (((𝐶 + 𝐴) + (𝐶 − 𝐴)) / 2) = ((2 · 𝐶) / 2)) |
23 | | 2cn 10968 |
. . . 4
⊢ 2 ∈
ℂ |
24 | | 2ne0 10990 |
. . . 4
⊢ 2 ≠
0 |
25 | | divcan3 10590 |
. . . 4
⊢ ((𝐶 ∈ ℂ ∧ 2 ∈
ℂ ∧ 2 ≠ 0) → ((2 · 𝐶) / 2) = 𝐶) |
26 | 23, 24, 25 | mp3an23 1408 |
. . 3
⊢ (𝐶 ∈ ℂ → ((2
· 𝐶) / 2) = 𝐶) |
27 | 8, 26 | syl 17 |
. 2
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ((2 · 𝐶) / 2) = 𝐶) |
28 | 18, 22, 27 | 3eqtrrd 2649 |
1
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐶 = ((𝑀↑2) + (𝑁↑2))) |