Metamath Proof Explorer


Theorem divgcdodd

Description: Either A / ( A gcd B ) is odd or B / ( A gcd B ) is odd. (Contributed by Scott Fenton, 19-Apr-2014)

Ref Expression
Assertion divgcdodd ⊢ A ∈ ℕ ∧ B ∈ ℕ → ¬ 2 ∥ A A gcd B ∨ ¬ 2 ∥ B A gcd B

Proof

Step Hyp Ref Expression
1 n2dvds1 ⊢ ¬ 2 ∥ 1
2 2z ⊢ 2 ∈ ℤ
3 nnz ⊢ A ∈ ℕ → A ∈ ℤ
4 nnz ⊢ B ∈ ℕ → B ∈ ℤ
5 gcddvds ⊢ A ∈ ℤ ∧ B ∈ ℤ → A gcd B ∥ A ∧ A gcd B ∥ B
6 3 4 5 syl2an ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B ∥ A ∧ A gcd B ∥ B
7 6 simpld ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B ∥ A
8 gcdnncl ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B ∈ ℕ
9 8 nnzd ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B ∈ ℤ
10 8 nnne0d ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B ≠ 0
11 3 adantr ⊢ A ∈ ℕ ∧ B ∈ ℕ → A ∈ ℤ
12 dvdsval2 ⊢ A gcd B ∈ ℤ ∧ A gcd B ≠ 0 ∧ A ∈ ℤ → A gcd B ∥ A ↔ A A gcd B ∈ ℤ
13 9 10 11 12 syl3anc ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B ∥ A ↔ A A gcd B ∈ ℤ
14 7 13 mpbid ⊢ A ∈ ℕ ∧ B ∈ ℕ → A A gcd B ∈ ℤ
15 6 simprd ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B ∥ B
16 4 adantl ⊢ A ∈ ℕ ∧ B ∈ ℕ → B ∈ ℤ
17 dvdsval2 ⊢ A gcd B ∈ ℤ ∧ A gcd B ≠ 0 ∧ B ∈ ℤ → A gcd B ∥ B ↔ B A gcd B ∈ ℤ
18 9 10 16 17 syl3anc ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B ∥ B ↔ B A gcd B ∈ ℤ
19 15 18 mpbid ⊢ A ∈ ℕ ∧ B ∈ ℕ → B A gcd B ∈ ℤ
20 dvdsgcdb ⊢ 2 ∈ ℤ ∧ A A gcd B ∈ ℤ ∧ B A gcd B ∈ ℤ → 2 ∥ A A gcd B ∧ 2 ∥ B A gcd B ↔ 2 ∥ A A gcd B gcd B A gcd B
21 2 14 19 20 mp3an2i ⊢ A ∈ ℕ ∧ B ∈ ℕ → 2 ∥ A A gcd B ∧ 2 ∥ B A gcd B ↔ 2 ∥ A A gcd B gcd B A gcd B
22 gcddiv ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ A gcd B ∈ ℕ ∧ A gcd B ∥ A ∧ A gcd B ∥ B → A gcd B A gcd B = A A gcd B gcd B A gcd B
23 11 16 8 6 22 syl31anc ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B A gcd B = A A gcd B gcd B A gcd B
24 8 nncnd ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B ∈ ℂ
25 24 10 dividd ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B A gcd B = 1
26 23 25 eqtr3d ⊢ A ∈ ℕ ∧ B ∈ ℕ → A A gcd B gcd B A gcd B = 1
27 26 breq2d ⊢ A ∈ ℕ ∧ B ∈ ℕ → 2 ∥ A A gcd B gcd B A gcd B ↔ 2 ∥ 1
28 27 biimpd ⊢ A ∈ ℕ ∧ B ∈ ℕ → 2 ∥ A A gcd B gcd B A gcd B → 2 ∥ 1
29 21 28 sylbid ⊢ A ∈ ℕ ∧ B ∈ ℕ → 2 ∥ A A gcd B ∧ 2 ∥ B A gcd B → 2 ∥ 1
30 29 expdimp ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ 2 ∥ A A gcd B → 2 ∥ B A gcd B → 2 ∥ 1
31 1 30 mtoi ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ 2 ∥ A A gcd B → ¬ 2 ∥ B A gcd B
32 31 ex ⊢ A ∈ ℕ ∧ B ∈ ℕ → 2 ∥ A A gcd B → ¬ 2 ∥ B A gcd B
33 imor ⊢ 2 ∥ A A gcd B → ¬ 2 ∥ B A gcd B ↔ ¬ 2 ∥ A A gcd B ∨ ¬ 2 ∥ B A gcd B
34 32 33 sylib ⊢ A ∈ ℕ ∧ B ∈ ℕ → ¬ 2 ∥ A A gcd B ∨ ¬ 2 ∥ B A gcd B