Metamath Proof Explorer


Theorem pythagtriplem4

Description: Lemma for pythagtrip . Show that C - B and C + B are relatively prime. (Contributed by Scott Fenton, 12-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion pythagtriplem4 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B = 1

Proof

Step Hyp Ref Expression
1 simp3r ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → ¬ 2 ∥ A
2 nnz ⊢ C ∈ ℕ → C ∈ ℤ
3 nnz ⊢ B ∈ ℕ → B ∈ ℤ
4 zsubcl ⊢ C ∈ ℤ ∧ B ∈ ℤ → C − B ∈ ℤ
5 2 3 4 syl2anr ⊢ B ∈ ℕ ∧ C ∈ ℕ → C − B ∈ ℤ
6 5 3adant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → C − B ∈ ℤ
7 6 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B ∈ ℤ
8 simp13 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C ∈ ℕ
9 simp12 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → B ∈ ℕ
10 8 9 nnaddcld ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B ∈ ℕ
11 10 nnzd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B ∈ ℤ
12 gcddvds ⊢ C − B ∈ ℤ ∧ C + B ∈ ℤ → C − B gcd C + B ∥ C − B ∧ C − B gcd C + B ∥ C + B
13 7 11 12 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B ∥ C − B ∧ C − B gcd C + B ∥ C + B
14 13 simprd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B ∥ C + B
15 breq1 ⊢ C − B gcd C + B = 2 → C − B gcd C + B ∥ C + B ↔ 2 ∥ C + B
16 15 biimpd ⊢ C − B gcd C + B = 2 → C − B gcd C + B ∥ C + B → 2 ∥ C + B
17 14 16 mpan9 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → 2 ∥ C + B
18 2z ⊢ 2 ∈ ℤ
19 simpl13 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → C ∈ ℕ
20 19 nnzd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → C ∈ ℤ
21 simpl12 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → B ∈ ℕ
22 21 nnzd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → B ∈ ℤ
23 20 22 zaddcld ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → C + B ∈ ℤ
24 20 22 zsubcld ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → C − B ∈ ℤ
25 dvdsmultr1 ⊢ 2 ∈ ℤ ∧ C + B ∈ ℤ ∧ C − B ∈ ℤ → 2 ∥ C + B → 2 ∥ C + B ⁢ C − B
26 18 23 24 25 mp3an2i ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → 2 ∥ C + B → 2 ∥ C + B ⁢ C − B
27 17 26 mpd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → 2 ∥ C + B ⁢ C − B
28 19 nncnd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → C ∈ ℂ
29 21 nncnd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → B ∈ ℂ
30 subsq ⊢ C ∈ ℂ ∧ B ∈ ℂ → C 2 − B 2 = C + B ⁢ C − B
31 28 29 30 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → C 2 − B 2 = C + B ⁢ C − B
32 27 31 breqtrrd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → 2 ∥ C 2 − B 2
33 simpl2 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → A 2 + B 2 = C 2
34 33 oveq1d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → A 2 + B 2 - B 2 = C 2 − B 2
35 simpl11 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → A ∈ ℕ
36 35 nnsqcld ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → A 2 ∈ ℕ
37 36 nncnd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → A 2 ∈ ℂ
38 21 nnsqcld ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → B 2 ∈ ℕ
39 38 nncnd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → B 2 ∈ ℂ
40 37 39 pncand ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → A 2 + B 2 - B 2 = A 2
41 34 40 eqtr3d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → C 2 − B 2 = A 2
42 32 41 breqtrd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → 2 ∥ A 2
43 nnz ⊢ A ∈ ℕ → A ∈ ℤ
44 43 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → A ∈ ℤ
45 44 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → A ∈ ℤ
46 45 adantr ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → A ∈ ℤ
47 2prm ⊢ 2 ∈ ℙ
48 2nn ⊢ 2 ∈ ℕ
49 prmdvdsexp ⊢ 2 ∈ ℙ ∧ A ∈ ℤ ∧ 2 ∈ ℕ → 2 ∥ A 2 ↔ 2 ∥ A
50 47 48 49 mp3an13 ⊢ A ∈ ℤ → 2 ∥ A 2 ↔ 2 ∥ A
51 46 50 syl ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → 2 ∥ A 2 ↔ 2 ∥ A
52 42 51 mpbid ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A ∧ C − B gcd C + B = 2 → 2 ∥ A
53 1 52 mtand ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → ¬ C − B gcd C + B = 2
54 neg1z ⊢ − 1 ∈ ℤ
55 gcdaddm ⊢ − 1 ∈ ℤ ∧ C − B ∈ ℤ ∧ C + B ∈ ℤ → C − B gcd C + B = C − B gcd C + B + -1 ⁢ C − B
56 54 7 11 55 mp3an2i ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B = C − B gcd C + B + -1 ⁢ C − B
57 8 nncnd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C ∈ ℂ
58 9 nncnd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → B ∈ ℂ
59 pnncan ⊢ C ∈ ℂ ∧ B ∈ ℂ ∧ B ∈ ℂ → C + B - C − B = B + B
60 59 3anidm23 ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B - C − B = B + B
61 subcl ⊢ C ∈ ℂ ∧ B ∈ ℂ → C − B ∈ ℂ
62 61 mulm1d ⊢ C ∈ ℂ ∧ B ∈ ℂ → -1 ⁢ C − B = − C − B
63 62 oveq2d ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B + -1 ⁢ C − B = C + B + − C − B
64 addcl ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B ∈ ℂ
65 64 61 negsubd ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B + − C − B = C + B - C − B
66 63 65 eqtrd ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B + -1 ⁢ C − B = C + B - C − B
67 2times ⊢ B ∈ ℂ → 2 ⁢ B = B + B
68 67 adantl ⊢ C ∈ ℂ ∧ B ∈ ℂ → 2 ⁢ B = B + B
69 60 66 68 3eqtr4d ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B + -1 ⁢ C − B = 2 ⁢ B
70 69 oveq2d ⊢ C ∈ ℂ ∧ B ∈ ℂ → C − B gcd C + B + -1 ⁢ C − B = C − B gcd 2 ⁢ B
71 57 58 70 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B + -1 ⁢ C − B = C − B gcd 2 ⁢ B
72 56 71 eqtrd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B = C − B gcd 2 ⁢ B
73 9 nnzd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → B ∈ ℤ
74 zmulcl ⊢ 2 ∈ ℤ ∧ B ∈ ℤ → 2 ⁢ B ∈ ℤ
75 18 73 74 sylancr ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 2 ⁢ B ∈ ℤ
76 gcddvds ⊢ C − B ∈ ℤ ∧ 2 ⁢ B ∈ ℤ → C − B gcd 2 ⁢ B ∥ C − B ∧ C − B gcd 2 ⁢ B ∥ 2 ⁢ B
77 7 75 76 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd 2 ⁢ B ∥ C − B ∧ C − B gcd 2 ⁢ B ∥ 2 ⁢ B
78 77 simprd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd 2 ⁢ B ∥ 2 ⁢ B
79 72 78 eqbrtrd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B ∥ 2 ⁢ B
80 1z ⊢ 1 ∈ ℤ
81 gcdaddm ⊢ 1 ∈ ℤ ∧ C − B ∈ ℤ ∧ C + B ∈ ℤ → C − B gcd C + B = C − B gcd C + B + 1 ⁢ C − B
82 80 7 11 81 mp3an2i ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B = C − B gcd C + B + 1 ⁢ C − B
83 ppncan ⊢ C ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → C + B + C − B = C + C
84 83 3anidm13 ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B + C − B = C + C
85 61 mullidd ⊢ C ∈ ℂ ∧ B ∈ ℂ → 1 ⁢ C − B = C − B
86 85 oveq2d ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B + 1 ⁢ C − B = C + B + C − B
87 2times ⊢ C ∈ ℂ → 2 ⁢ C = C + C
88 87 adantr ⊢ C ∈ ℂ ∧ B ∈ ℂ → 2 ⁢ C = C + C
89 84 86 88 3eqtr4d ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B + 1 ⁢ C − B = 2 ⁢ C
90 57 58 89 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B + 1 ⁢ C − B = 2 ⁢ C
91 90 oveq2d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B + 1 ⁢ C − B = C − B gcd 2 ⁢ C
92 82 91 eqtrd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B = C − B gcd 2 ⁢ C
93 8 nnzd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C ∈ ℤ
94 zmulcl ⊢ 2 ∈ ℤ ∧ C ∈ ℤ → 2 ⁢ C ∈ ℤ
95 18 93 94 sylancr ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 2 ⁢ C ∈ ℤ
96 gcddvds ⊢ C − B ∈ ℤ ∧ 2 ⁢ C ∈ ℤ → C − B gcd 2 ⁢ C ∥ C − B ∧ C − B gcd 2 ⁢ C ∥ 2 ⁢ C
97 7 95 96 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd 2 ⁢ C ∥ C − B ∧ C − B gcd 2 ⁢ C ∥ 2 ⁢ C
98 97 simprd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd 2 ⁢ C ∥ 2 ⁢ C
99 92 98 eqbrtrd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B ∥ 2 ⁢ C
100 nnaddcl ⊢ C ∈ ℕ ∧ B ∈ ℕ → C + B ∈ ℕ
101 100 nnne0d ⊢ C ∈ ℕ ∧ B ∈ ℕ → C + B ≠ 0
102 101 ancoms ⊢ B ∈ ℕ ∧ C ∈ ℕ → C + B ≠ 0
103 102 3adant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → C + B ≠ 0
104 103 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B ≠ 0
105 104 neneqd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → ¬ C + B = 0
106 105 intnand ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → ¬ C − B = 0 ∧ C + B = 0
107 gcdn0cl ⊢ C − B ∈ ℤ ∧ C + B ∈ ℤ ∧ ¬ C − B = 0 ∧ C + B = 0 → C − B gcd C + B ∈ ℕ
108 7 11 106 107 syl21anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B ∈ ℕ
109 108 nnzd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B ∈ ℤ
110 dvdsgcd ⊢ C − B gcd C + B ∈ ℤ ∧ 2 ⁢ B ∈ ℤ ∧ 2 ⁢ C ∈ ℤ → C − B gcd C + B ∥ 2 ⁢ B ∧ C − B gcd C + B ∥ 2 ⁢ C → C − B gcd C + B ∥ 2 ⁢ B gcd 2 ⁢ C
111 109 75 95 110 syl3anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B ∥ 2 ⁢ B ∧ C − B gcd C + B ∥ 2 ⁢ C → C − B gcd C + B ∥ 2 ⁢ B gcd 2 ⁢ C
112 79 99 111 mp2and ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B ∥ 2 ⁢ B gcd 2 ⁢ C
113 2nn0 ⊢ 2 ∈ ℕ 0
114 mulgcd ⊢ 2 ∈ ℕ 0 ∧ B ∈ ℤ ∧ C ∈ ℤ → 2 ⁢ B gcd 2 ⁢ C = 2 ⁢ B gcd C
115 113 73 93 114 mp3an2i ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 2 ⁢ B gcd 2 ⁢ C = 2 ⁢ B gcd C
116 pythagtriplem3 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → B gcd C = 1
117 116 oveq2d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 2 ⁢ B gcd C = 2 ⋅ 1
118 2t1e2 ⊢ 2 ⋅ 1 = 2
119 117 118 eqtrdi ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 2 ⁢ B gcd C = 2
120 115 119 eqtrd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 2 ⁢ B gcd 2 ⁢ C = 2
121 112 120 breqtrd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B ∥ 2
122 dvdsprime ⊢ 2 ∈ ℙ ∧ C − B gcd C + B ∈ ℕ → C − B gcd C + B ∥ 2 ↔ C − B gcd C + B = 2 ∨ C − B gcd C + B = 1
123 47 108 122 sylancr ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B ∥ 2 ↔ C − B gcd C + B = 2 ∨ C − B gcd C + B = 1
124 121 123 mpbid ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B = 2 ∨ C − B gcd C + B = 1
125 orel1 ⊢ ¬ C − B gcd C + B = 2 → C − B gcd C + B = 2 ∨ C − B gcd C + B = 1 → C − B gcd C + B = 1
126 53 124 125 sylc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B gcd C + B = 1