Metamath Proof Explorer


Theorem gausslemma2dlem1

Description: Lemma 1 for gausslemma2d . (Contributed by AV, 5-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
Assertion gausslemma2dlem1 ⊢ φ → H ! = ∏ k = 1 H R ⁡ k

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 1 2 gausslemma2dlem0b ⊢ φ → H ∈ ℕ
5 4 nnnn0d ⊢ φ → H ∈ ℕ 0
6 fprodfac ⊢ H ∈ ℕ 0 → H ! = ∏ l = 1 H l
7 5 6 syl ⊢ φ → H ! = ∏ l = 1 H l
8 id ⊢ l = R ⁡ k → l = R ⁡ k
9 fzfid ⊢ φ → 1 … H ∈ Fin
10 fzfi ⊢ 1 … H ∈ Fin
11 ovex ⊢ x ⋅ 2 ∈ V
12 ovex ⊢ P − x ⋅ 2 ∈ V
13 11 12 ifex ⊢ if x ⋅ 2 < P 2 x ⋅ 2 P − x ⋅ 2 ∈ V
14 13 3 fnmpti ⊢ R Fn 1 … H
15 1 2 3 gausslemma2dlem1a ⊢ φ → ran ⁡ R = 1 … H
16 rneqdmfinf1o ⊢ 1 … H ∈ Fin ∧ R Fn 1 … H ∧ ran ⁡ R = 1 … H → R : 1 … H ⟶ 1-1 onto 1 … H
17 10 14 15 16 mp3an12i ⊢ φ → R : 1 … H ⟶ 1-1 onto 1 … H
18 eqidd ⊢ φ ∧ k ∈ 1 … H → R ⁡ k = R ⁡ k
19 elfzelz ⊢ l ∈ 1 … H → l ∈ ℤ
20 19 zcnd ⊢ l ∈ 1 … H → l ∈ ℂ
21 20 adantl ⊢ φ ∧ l ∈ 1 … H → l ∈ ℂ
22 8 9 17 18 21 fprodf1o ⊢ φ → ∏ l = 1 H l = ∏ k = 1 H R ⁡ k
23 7 22 eqtrd ⊢ φ → H ! = ∏ k = 1 H R ⁡ k