Metamath Proof Explorer


Theorem gausslemma2dlem7

Description: Lemma 7 for gausslemma2d . (Contributed by AV, 13-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
gausslemma2d.n ⊢ N = H − M
Assertion gausslemma2dlem7 ⊢ φ → − 1 N ⁢ 2 H mod P = 1

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 gausslemma2d.n ⊢ N = H − M
6 1 2 3 4 5 gausslemma2dlem6 ⊢ φ → H ! mod P = − 1 N ⁢ 2 H ⁢ H ! mod P
7 1 2 gausslemma2dlem0b ⊢ φ → H ∈ ℕ
8 7 nnnn0d ⊢ φ → H ∈ ℕ 0
9 8 faccld ⊢ φ → H ! ∈ ℕ
10 9 nncnd ⊢ φ → H ! ∈ ℂ
11 10 mullidd ⊢ φ → 1 ⁢ H ! = H !
12 11 eqcomd ⊢ φ → H ! = 1 ⁢ H !
13 12 oveq1d ⊢ φ → H ! mod P = 1 ⁢ H ! mod P
14 13 eqeq1d ⊢ φ → H ! mod P = − 1 N ⁢ 2 H ⁢ H ! mod P ↔ 1 ⁢ H ! mod P = − 1 N ⁢ 2 H ⁢ H ! mod P
15 1zzd ⊢ φ → 1 ∈ ℤ
16 neg1z ⊢ − 1 ∈ ℤ
17 1 4 2 5 gausslemma2dlem0h ⊢ φ → N ∈ ℕ 0
18 zexpcl ⊢ − 1 ∈ ℤ ∧ N ∈ ℕ 0 → − 1 N ∈ ℤ
19 16 17 18 sylancr ⊢ φ → − 1 N ∈ ℤ
20 2z ⊢ 2 ∈ ℤ
21 zexpcl ⊢ 2 ∈ ℤ ∧ H ∈ ℕ 0 → 2 H ∈ ℤ
22 20 8 21 sylancr ⊢ φ → 2 H ∈ ℤ
23 19 22 zmulcld ⊢ φ → − 1 N ⁢ 2 H ∈ ℤ
24 9 nnzd ⊢ φ → H ! ∈ ℤ
25 eldifi ⊢ P ∈ ℙ ∖ 2 → P ∈ ℙ
26 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
27 1 25 26 3syl ⊢ φ → P ∈ ℕ
28 1 2 gausslemma2dlem0c ⊢ φ → H ! gcd P = 1
29 cncongrcoprm ⊢ 1 ∈ ℤ ∧ − 1 N ⁢ 2 H ∈ ℤ ∧ H ! ∈ ℤ ∧ P ∈ ℕ ∧ H ! gcd P = 1 → 1 ⁢ H ! mod P = − 1 N ⁢ 2 H ⁢ H ! mod P ↔ 1 mod P = − 1 N ⁢ 2 H mod P
30 15 23 24 27 28 29 syl32anc ⊢ φ → 1 ⁢ H ! mod P = − 1 N ⁢ 2 H ⁢ H ! mod P ↔ 1 mod P = − 1 N ⁢ 2 H mod P
31 14 30 bitrd ⊢ φ → H ! mod P = − 1 N ⁢ 2 H ⁢ H ! mod P ↔ 1 mod P = − 1 N ⁢ 2 H mod P
32 simpr ⊢ φ ∧ 1 mod P = − 1 N ⁢ 2 H mod P → 1 mod P = − 1 N ⁢ 2 H mod P
33 26 nnred ⊢ P ∈ ℙ → P ∈ ℝ
34 prmgt1 ⊢ P ∈ ℙ → 1 < P
35 33 34 jca ⊢ P ∈ ℙ → P ∈ ℝ ∧ 1 < P
36 1mod ⊢ P ∈ ℝ ∧ 1 < P → 1 mod P = 1
37 1 25 35 36 4syl ⊢ φ → 1 mod P = 1
38 37 adantr ⊢ φ ∧ 1 mod P = − 1 N ⁢ 2 H mod P → 1 mod P = 1
39 32 38 eqtr3d ⊢ φ ∧ 1 mod P = − 1 N ⁢ 2 H mod P → − 1 N ⁢ 2 H mod P = 1
40 39 ex ⊢ φ → 1 mod P = − 1 N ⁢ 2 H mod P → − 1 N ⁢ 2 H mod P = 1
41 31 40 sylbid ⊢ φ → H ! mod P = − 1 N ⁢ 2 H ⁢ H ! mod P → − 1 N ⁢ 2 H mod P = 1
42 6 41 mpd ⊢ φ → − 1 N ⁢ 2 H mod P = 1