Metamath Proof Explorer


Theorem gausslemma2dlem5a

Description: Lemma for gausslemma2dlem5 . (Contributed by AV, 8-Jul-2021)

Ref Expression
Hypotheses gausslemma2d.p ⊢ φ → P ∈ ℙ ∖ 2
gausslemma2d.h ⊢ H = P − 1 2
gausslemma2d.r ⊢ R = x ∈ 1 … H ⟼ if x ⋅ 2 < P 2 x ⋅ 2 P − x ⋅ 2
gausslemma2d.m ⊢ M = P 4
Assertion gausslemma2dlem5a ⊢ φ → ∏ k = M + 1 H R ⁡ k mod P = ∏ k = M + 1 H -1 ⁢ k ⋅ 2 mod P

Proof

Step Hyp Ref Expression
1 gausslemma2d.p ⊢ φ → P ∈ ℙ ∖ 2
2 gausslemma2d.h ⊢ H = P − 1 2
3 gausslemma2d.r ⊢ R = x ∈ 1 … H ⟼ if x ⋅ 2 < P 2 x ⋅ 2 P − x ⋅ 2
4 gausslemma2d.m ⊢ M = P 4
5 1 2 3 4 gausslemma2dlem3 ⊢ φ → ∀ k ∈ M + 1 … H R ⁡ k = P − k ⋅ 2
6 prodeq2 ⊢ ∀ k ∈ M + 1 … H R ⁡ k = P − k ⋅ 2 → ∏ k = M + 1 H R ⁡ k = ∏ k = M + 1 H P − k ⋅ 2
7 6 oveq1d ⊢ ∀ k ∈ M + 1 … H R ⁡ k = P − k ⋅ 2 → ∏ k = M + 1 H R ⁡ k mod P = ∏ k = M + 1 H P − k ⋅ 2 mod P
8 5 7 syl ⊢ φ → ∏ k = M + 1 H R ⁡ k mod P = ∏ k = M + 1 H P − k ⋅ 2 mod P
9 eldifi ⊢ P ∈ ℙ ∖ 2 → P ∈ ℙ
10 fzfid ⊢ P ∈ ℙ → M + 1 … H ∈ Fin
11 prmz ⊢ P ∈ ℙ → P ∈ ℤ
12 11 adantr ⊢ P ∈ ℙ ∧ k ∈ M + 1 … H → P ∈ ℤ
13 elfzelz ⊢ k ∈ M + 1 … H → k ∈ ℤ
14 2z ⊢ 2 ∈ ℤ
15 14 a1i ⊢ k ∈ M + 1 … H → 2 ∈ ℤ
16 13 15 zmulcld ⊢ k ∈ M + 1 … H → k ⋅ 2 ∈ ℤ
17 16 adantl ⊢ P ∈ ℙ ∧ k ∈ M + 1 … H → k ⋅ 2 ∈ ℤ
18 12 17 zsubcld ⊢ P ∈ ℙ ∧ k ∈ M + 1 … H → P − k ⋅ 2 ∈ ℤ
19 neg1z ⊢ − 1 ∈ ℤ
20 19 a1i ⊢ k ∈ M + 1 … H → − 1 ∈ ℤ
21 20 16 zmulcld ⊢ k ∈ M + 1 … H → -1 ⁢ k ⋅ 2 ∈ ℤ
22 21 adantl ⊢ P ∈ ℙ ∧ k ∈ M + 1 … H → -1 ⁢ k ⋅ 2 ∈ ℤ
23 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
24 16 zcnd ⊢ k ∈ M + 1 … H → k ⋅ 2 ∈ ℂ
25 24 mulm1d ⊢ k ∈ M + 1 … H → -1 ⁢ k ⋅ 2 = − k ⋅ 2
26 25 adantl ⊢ P ∈ ℙ ∧ k ∈ M + 1 … H → -1 ⁢ k ⋅ 2 = − k ⋅ 2
27 26 oveq1d ⊢ P ∈ ℙ ∧ k ∈ M + 1 … H → -1 ⁢ k ⋅ 2 mod P = − k ⋅ 2 mod P
28 16 zred ⊢ k ∈ M + 1 … H → k ⋅ 2 ∈ ℝ
29 23 nnrpd ⊢ P ∈ ℙ → P ∈ ℝ +
30 negmod ⊢ k ⋅ 2 ∈ ℝ ∧ P ∈ ℝ + → − k ⋅ 2 mod P = P − k ⋅ 2 mod P
31 28 29 30 syl2anr ⊢ P ∈ ℙ ∧ k ∈ M + 1 … H → − k ⋅ 2 mod P = P − k ⋅ 2 mod P
32 27 31 eqtr2d ⊢ P ∈ ℙ ∧ k ∈ M + 1 … H → P − k ⋅ 2 mod P = -1 ⁢ k ⋅ 2 mod P
33 10 18 22 23 32 fprodmodd ⊢ P ∈ ℙ → ∏ k = M + 1 H P − k ⋅ 2 mod P = ∏ k = M + 1 H -1 ⁢ k ⋅ 2 mod P
34 1 9 33 3syl ⊢ φ → ∏ k = M + 1 H P − k ⋅ 2 mod P = ∏ k = M + 1 H -1 ⁢ k ⋅ 2 mod P
35 8 34 eqtrd ⊢ φ → ∏ k = M + 1 H R ⁡ k mod P = ∏ k = M + 1 H -1 ⁢ k ⋅ 2 mod P