Metamath Proof Explorer


Theorem knoppndvlem16

Description: Lemma for knoppndv . (Contributed by Asger C. Ipsen, 19-Jul-2021)

Ref Expression
Hypotheses knoppndvlem16.a ⊢ A = 2 ⋅ N − J 2 ⋅ M
knoppndvlem16.b ⊢ B = 2 ⋅ N − J 2 ⁢ M + 1
knoppndvlem16.j ⊢ φ → J ∈ ℕ 0
knoppndvlem16.m ⊢ φ → M ∈ ℤ
knoppndvlem16.n ⊢ φ → N ∈ ℕ
Assertion knoppndvlem16 ⊢ φ → B − A = 2 ⋅ N − J 2

Proof

Step Hyp Ref Expression
1 knoppndvlem16.a ⊢ A = 2 ⋅ N − J 2 ⋅ M
2 knoppndvlem16.b ⊢ B = 2 ⋅ N − J 2 ⁢ M + 1
3 knoppndvlem16.j ⊢ φ → J ∈ ℕ 0
4 knoppndvlem16.m ⊢ φ → M ∈ ℤ
5 knoppndvlem16.n ⊢ φ → N ∈ ℕ
6 2 a1i ⊢ φ → B = 2 ⋅ N − J 2 ⁢ M + 1
7 1 a1i ⊢ φ → A = 2 ⋅ N − J 2 ⋅ M
8 6 7 oveq12d ⊢ φ → B − A = 2 ⋅ N − J 2 ⁢ M + 1 − 2 ⋅ N − J 2 ⋅ M
9 2cnd ⊢ φ → 2 ∈ ℂ
10 5 nncnd ⊢ φ → N ∈ ℂ
11 9 10 mulcld ⊢ φ → 2 ⋅ N ∈ ℂ
12 2ne0 ⊢ 2 ≠ 0
13 12 a1i ⊢ φ → 2 ≠ 0
14 5 nnne0d ⊢ φ → N ≠ 0
15 9 10 13 14 mulne0d ⊢ φ → 2 ⋅ N ≠ 0
16 3 nn0zd ⊢ φ → J ∈ ℤ
17 16 znegcld ⊢ φ → − J ∈ ℤ
18 11 15 17 expclzd ⊢ φ → 2 ⋅ N − J ∈ ℂ
19 9 10 15 mulne0bad ⊢ φ → 2 ≠ 0
20 18 9 19 divcld ⊢ φ → 2 ⋅ N − J 2 ∈ ℂ
21 4 zcnd ⊢ φ → M ∈ ℂ
22 1cnd ⊢ φ → 1 ∈ ℂ
23 21 22 addcld ⊢ φ → M + 1 ∈ ℂ
24 20 23 21 subdid ⊢ φ → 2 ⋅ N − J 2 ⁢ M + 1 - M = 2 ⋅ N − J 2 ⁢ M + 1 − 2 ⋅ N − J 2 ⋅ M
25 24 eqcomd ⊢ φ → 2 ⋅ N − J 2 ⁢ M + 1 − 2 ⋅ N − J 2 ⋅ M = 2 ⋅ N − J 2 ⁢ M + 1 - M
26 21 22 pncan2d ⊢ φ → M + 1 - M = 1
27 26 oveq2d ⊢ φ → 2 ⋅ N − J 2 ⁢ M + 1 - M = 2 ⋅ N − J 2 ⋅ 1
28 20 mulridd ⊢ φ → 2 ⋅ N − J 2 ⋅ 1 = 2 ⋅ N − J 2
29 27 28 eqtrd ⊢ φ → 2 ⋅ N − J 2 ⁢ M + 1 - M = 2 ⋅ N − J 2
30 8 25 29 3eqtrd ⊢ φ → B − A = 2 ⋅ N − J 2