Metamath Proof Explorer


Theorem omoe

Description: The difference of two odds is even. (Contributed by Scott Fenton, 7-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion omoe ⊢ A ∈ ℤ ∧ ¬ 2 ∥ A ∧ B ∈ ℤ ∧ ¬ 2 ∥ B → 2 ∥ A − B

Proof

Step Hyp Ref Expression
1 odd2np1 ⊢ A ∈ ℤ → ¬ 2 ∥ A ↔ ∃ a ∈ ℤ 2 ⁢ a + 1 = A
2 odd2np1 ⊢ B ∈ ℤ → ¬ 2 ∥ B ↔ ∃ b ∈ ℤ 2 ⁢ b + 1 = B
3 1 2 bi2anan9 ⊢ A ∈ ℤ ∧ B ∈ ℤ → ¬ 2 ∥ A ∧ ¬ 2 ∥ B ↔ ∃ a ∈ ℤ 2 ⁢ a + 1 = A ∧ ∃ b ∈ ℤ 2 ⁢ b + 1 = B
4 reeanv ⊢ ∃ a ∈ ℤ ∃ b ∈ ℤ 2 ⁢ a + 1 = A ∧ 2 ⁢ b + 1 = B ↔ ∃ a ∈ ℤ 2 ⁢ a + 1 = A ∧ ∃ b ∈ ℤ 2 ⁢ b + 1 = B
5 2z ⊢ 2 ∈ ℤ
6 zsubcl ⊢ a ∈ ℤ ∧ b ∈ ℤ → a − b ∈ ℤ
7 dvdsmul1 ⊢ 2 ∈ ℤ ∧ a − b ∈ ℤ → 2 ∥ 2 ⁢ a − b
8 5 6 7 sylancr ⊢ a ∈ ℤ ∧ b ∈ ℤ → 2 ∥ 2 ⁢ a − b
9 zcn ⊢ a ∈ ℤ → a ∈ ℂ
10 zcn ⊢ b ∈ ℤ → b ∈ ℂ
11 2cn ⊢ 2 ∈ ℂ
12 mulcl ⊢ 2 ∈ ℂ ∧ a ∈ ℂ → 2 ⁢ a ∈ ℂ
13 11 12 mpan ⊢ a ∈ ℂ → 2 ⁢ a ∈ ℂ
14 mulcl ⊢ 2 ∈ ℂ ∧ b ∈ ℂ → 2 ⁢ b ∈ ℂ
15 11 14 mpan ⊢ b ∈ ℂ → 2 ⁢ b ∈ ℂ
16 ax-1cn ⊢ 1 ∈ ℂ
17 pnpcan2 ⊢ 2 ⁢ a ∈ ℂ ∧ 2 ⁢ b ∈ ℂ ∧ 1 ∈ ℂ → 2 ⁢ a + 1 - 2 ⁢ b + 1 = 2 ⁢ a − 2 ⁢ b
18 16 17 mp3an3 ⊢ 2 ⁢ a ∈ ℂ ∧ 2 ⁢ b ∈ ℂ → 2 ⁢ a + 1 - 2 ⁢ b + 1 = 2 ⁢ a − 2 ⁢ b
19 13 15 18 syl2an ⊢ a ∈ ℂ ∧ b ∈ ℂ → 2 ⁢ a + 1 - 2 ⁢ b + 1 = 2 ⁢ a − 2 ⁢ b
20 subdi ⊢ 2 ∈ ℂ ∧ a ∈ ℂ ∧ b ∈ ℂ → 2 ⁢ a − b = 2 ⁢ a − 2 ⁢ b
21 11 20 mp3an1 ⊢ a ∈ ℂ ∧ b ∈ ℂ → 2 ⁢ a − b = 2 ⁢ a − 2 ⁢ b
22 19 21 eqtr4d ⊢ a ∈ ℂ ∧ b ∈ ℂ → 2 ⁢ a + 1 - 2 ⁢ b + 1 = 2 ⁢ a − b
23 9 10 22 syl2an ⊢ a ∈ ℤ ∧ b ∈ ℤ → 2 ⁢ a + 1 - 2 ⁢ b + 1 = 2 ⁢ a − b
24 8 23 breqtrrd ⊢ a ∈ ℤ ∧ b ∈ ℤ → 2 ∥ 2 ⁢ a + 1 - 2 ⁢ b + 1
25 oveq12 ⊢ 2 ⁢ a + 1 = A ∧ 2 ⁢ b + 1 = B → 2 ⁢ a + 1 - 2 ⁢ b + 1 = A − B
26 25 breq2d ⊢ 2 ⁢ a + 1 = A ∧ 2 ⁢ b + 1 = B → 2 ∥ 2 ⁢ a + 1 - 2 ⁢ b + 1 ↔ 2 ∥ A − B
27 24 26 syl5ibcom ⊢ a ∈ ℤ ∧ b ∈ ℤ → 2 ⁢ a + 1 = A ∧ 2 ⁢ b + 1 = B → 2 ∥ A − B
28 27 rexlimivv ⊢ ∃ a ∈ ℤ ∃ b ∈ ℤ 2 ⁢ a + 1 = A ∧ 2 ⁢ b + 1 = B → 2 ∥ A − B
29 4 28 sylbir ⊢ ∃ a ∈ ℤ 2 ⁢ a + 1 = A ∧ ∃ b ∈ ℤ 2 ⁢ b + 1 = B → 2 ∥ A − B
30 3 29 biimtrdi ⊢ A ∈ ℤ ∧ B ∈ ℤ → ¬ 2 ∥ A ∧ ¬ 2 ∥ B → 2 ∥ A − B
31 30 imp ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ ¬ 2 ∥ A ∧ ¬ 2 ∥ B → 2 ∥ A − B
32 31 an4s ⊢ A ∈ ℤ ∧ ¬ 2 ∥ A ∧ B ∈ ℤ ∧ ¬ 2 ∥ B → 2 ∥ A − B