Metamath Proof Explorer


Theorem pythagtriplem2

Description: Lemma for pythagtrip . Prove the full version of one direction of the theorem. (Contributed by Scott Fenton, 28-Mar-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion pythagtriplem2 ⊢ A ∈ ℕ ∧ B ∈ ℕ → ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A B = k ⁢ m 2 − n 2 k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 → A 2 + B 2 = C 2

Proof

Step Hyp Ref Expression
1 ovex ⊢ k ⁢ m 2 − n 2 ∈ V
2 ovex ⊢ k ⁢ 2 ⁢ m ⁢ n ∈ V
3 preq12bg ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ k ⁢ m 2 − n 2 ∈ V ∧ k ⁢ 2 ⁢ m ⁢ n ∈ V → A B = k ⁢ m 2 − n 2 k ⁢ 2 ⁢ m ⁢ n ↔ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2
4 1 2 3 mpanr12 ⊢ A ∈ ℕ ∧ B ∈ ℕ → A B = k ⁢ m 2 − n 2 k ⁢ 2 ⁢ m ⁢ n ↔ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2
5 4 anbi1d ⊢ A ∈ ℕ ∧ B ∈ ℕ → A B = k ⁢ m 2 − n 2 k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ↔ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
6 andir ⊢ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
7 df-3an ⊢ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ↔ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2
8 df-3an ⊢ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
9 7 8 orbi12i ⊢ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
10 6 9 bitr4i ⊢ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
11 5 10 bitrdi ⊢ A ∈ ℕ ∧ B ∈ ℕ → A B = k ⁢ m 2 − n 2 k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ↔ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
12 11 rexbidv ⊢ A ∈ ℕ ∧ B ∈ ℕ → ∃ k ∈ ℕ A B = k ⁢ m 2 − n 2 k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ↔ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
13 12 2rexbidv ⊢ A ∈ ℕ ∧ B ∈ ℕ → ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A B = k ⁢ m 2 − n 2 k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ↔ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
14 r19.43 ⊢ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
15 14 2rexbii ⊢ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
16 r19.43 ⊢ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
17 16 rexbii ⊢ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
18 r19.43 ⊢ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
19 15 17 18 3bitri ⊢ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
20 13 19 bitrdi ⊢ A ∈ ℕ ∧ B ∈ ℕ → ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A B = k ⁢ m 2 − n 2 k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ↔ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2
21 pythagtriplem1 ⊢ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 → A 2 + B 2 = C 2
22 21 a1i ⊢ A ∈ ℕ ∧ B ∈ ℕ → ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 → A 2 + B 2 = C 2
23 3ancoma ⊢ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ B = k ⁢ m 2 − n 2 ∧ A = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2
24 23 rexbii ⊢ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ ∃ k ∈ ℕ B = k ⁢ m 2 − n 2 ∧ A = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2
25 24 2rexbii ⊢ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 ↔ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ B = k ⁢ m 2 − n 2 ∧ A = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2
26 pythagtriplem1 ⊢ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ B = k ⁢ m 2 − n 2 ∧ A = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 → B 2 + A 2 = C 2
27 25 26 sylbi ⊢ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 → B 2 + A 2 = C 2
28 nncn ⊢ A ∈ ℕ → A ∈ ℂ
29 28 sqcld ⊢ A ∈ ℕ → A 2 ∈ ℂ
30 nncn ⊢ B ∈ ℕ → B ∈ ℂ
31 30 sqcld ⊢ B ∈ ℕ → B 2 ∈ ℂ
32 addcom ⊢ A 2 ∈ ℂ ∧ B 2 ∈ ℂ → A 2 + B 2 = B 2 + A 2
33 29 31 32 syl2an ⊢ A ∈ ℕ ∧ B ∈ ℕ → A 2 + B 2 = B 2 + A 2
34 33 eqeq1d ⊢ A ∈ ℕ ∧ B ∈ ℕ → A 2 + B 2 = C 2 ↔ B 2 + A 2 = C 2
35 27 34 imbitrrid ⊢ A ∈ ℕ ∧ B ∈ ℕ → ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 → A 2 + B 2 = C 2
36 22 35 jaod ⊢ A ∈ ℕ ∧ B ∈ ℕ → ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ m 2 − n 2 ∧ B = k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 ∨ ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A = k ⁢ 2 ⁢ m ⁢ n ∧ B = k ⁢ m 2 − n 2 ∧ C = k ⁢ m 2 + n 2 → A 2 + B 2 = C 2
37 20 36 sylbid ⊢ A ∈ ℕ ∧ B ∈ ℕ → ∃ n ∈ ℕ ∃ m ∈ ℕ ∃ k ∈ ℕ A B = k ⁢ m 2 − n 2 k ⁢ 2 ⁢ m ⁢ n ∧ C = k ⁢ m 2 + n 2 → A 2 + B 2 = C 2