Metamath Proof Explorer


Theorem gausslemma2dlem0c

Description: Auxiliary lemma 3 for gausslemma2d . (Contributed by AV, 13-Jul-2021)

Ref Expression
Hypotheses gausslemma2dlem0a.p ⊢ φ → P ∈ ℙ ∖ 2
gausslemma2dlem0b.h ⊢ H = P − 1 2
Assertion gausslemma2dlem0c ⊢ φ → H ! gcd P = 1

Proof

Step Hyp Ref Expression
1 gausslemma2dlem0a.p ⊢ φ → P ∈ ℙ ∖ 2
2 gausslemma2dlem0b.h ⊢ H = P − 1 2
3 eldifi ⊢ P ∈ ℙ ∖ 2 → P ∈ ℙ
4 1 3 syl ⊢ φ → P ∈ ℙ
5 1 2 gausslemma2dlem0b ⊢ φ → H ∈ ℕ
6 5 nnnn0d ⊢ φ → H ∈ ℕ 0
7 4 6 jca ⊢ φ → P ∈ ℙ ∧ H ∈ ℕ 0
8 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
9 nnre ⊢ P ∈ ℕ → P ∈ ℝ
10 peano2rem ⊢ P ∈ ℝ → P − 1 ∈ ℝ
11 9 10 syl ⊢ P ∈ ℕ → P − 1 ∈ ℝ
12 2re ⊢ 2 ∈ ℝ
13 12 a1i ⊢ P ∈ ℕ → 2 ∈ ℝ
14 13 9 remulcld ⊢ P ∈ ℕ → 2 ⁢ P ∈ ℝ
15 9 ltm1d ⊢ P ∈ ℕ → P − 1 < P
16 nnnn0 ⊢ P ∈ ℕ → P ∈ ℕ 0
17 16 nn0ge0d ⊢ P ∈ ℕ → 0 ≤ P
18 1le2 ⊢ 1 ≤ 2
19 18 a1i ⊢ P ∈ ℕ → 1 ≤ 2
20 9 13 17 19 lemulge12d ⊢ P ∈ ℕ → P ≤ 2 ⁢ P
21 11 9 14 15 20 ltletrd ⊢ P ∈ ℕ → P − 1 < 2 ⁢ P
22 2pos ⊢ 0 < 2
23 12 22 pm3.2i ⊢ 2 ∈ ℝ ∧ 0 < 2
24 23 a1i ⊢ P ∈ ℕ → 2 ∈ ℝ ∧ 0 < 2
25 ltdivmul ⊢ P − 1 ∈ ℝ ∧ P ∈ ℝ ∧ 2 ∈ ℝ ∧ 0 < 2 → P − 1 2 < P ↔ P − 1 < 2 ⁢ P
26 11 9 24 25 syl3anc ⊢ P ∈ ℕ → P − 1 2 < P ↔ P − 1 < 2 ⁢ P
27 21 26 mpbird ⊢ P ∈ ℕ → P − 1 2 < P
28 1 3 8 27 4syl ⊢ φ → P − 1 2 < P
29 2 28 eqbrtrid ⊢ φ → H < P
30 prmndvdsfaclt ⊢ P ∈ ℙ ∧ H ∈ ℕ 0 → H < P → ¬ P ∥ H !
31 7 29 30 sylc ⊢ φ → ¬ P ∥ H !
32 6 faccld ⊢ φ → H ! ∈ ℕ
33 32 nnzd ⊢ φ → H ! ∈ ℤ
34 nnz ⊢ P ∈ ℕ → P ∈ ℤ
35 1 3 8 34 4syl ⊢ φ → P ∈ ℤ
36 33 35 gcdcomd ⊢ φ → H ! gcd P = P gcd H !
37 36 eqeq1d ⊢ φ → H ! gcd P = 1 ↔ P gcd H ! = 1
38 coprm ⊢ P ∈ ℙ ∧ H ! ∈ ℤ → ¬ P ∥ H ! ↔ P gcd H ! = 1
39 4 33 38 syl2anc ⊢ φ → ¬ P ∥ H ! ↔ P gcd H ! = 1
40 37 39 bitr4d ⊢ φ → H ! gcd P = 1 ↔ ¬ P ∥ H !
41 31 40 mpbird ⊢ φ → H ! gcd P = 1