Metamath Proof Explorer


Theorem gausslemma2dlem0f

Description: Auxiliary lemma 6 for gausslemma2d . (Contributed by AV, 9-Jul-2021)

Ref Expression
Hypotheses gausslemma2dlem0.p ⊢ φ → P ∈ ℙ ∖ 2
gausslemma2dlem0.m ⊢ M = P 4
gausslemma2dlem0.h ⊢ H = P − 1 2
Assertion gausslemma2dlem0f ⊢ φ → M + 1 ≤ H

Proof

Step Hyp Ref Expression
1 gausslemma2dlem0.p ⊢ φ → P ∈ ℙ ∖ 2
2 gausslemma2dlem0.m ⊢ M = P 4
3 gausslemma2dlem0.h ⊢ H = P − 1 2
4 eldifsn ⊢ P ∈ ℙ ∖ 2 ↔ P ∈ ℙ ∧ P ≠ 2
5 prm23ge5 ⊢ P ∈ ℙ → P = 2 ∨ P = 3 ∨ P ∈ ℤ ≥ 5
6 eqneqall ⊢ P = 2 → P ≠ 2 → P = 3 ∨ P ∈ ℤ ≥ 5
7 orc ⊢ P = 3 → P = 3 ∨ P ∈ ℤ ≥ 5
8 7 a1d ⊢ P = 3 → P ≠ 2 → P = 3 ∨ P ∈ ℤ ≥ 5
9 olc ⊢ P ∈ ℤ ≥ 5 → P = 3 ∨ P ∈ ℤ ≥ 5
10 9 a1d ⊢ P ∈ ℤ ≥ 5 → P ≠ 2 → P = 3 ∨ P ∈ ℤ ≥ 5
11 6 8 10 3jaoi ⊢ P = 2 ∨ P = 3 ∨ P ∈ ℤ ≥ 5 → P ≠ 2 → P = 3 ∨ P ∈ ℤ ≥ 5
12 5 11 syl ⊢ P ∈ ℙ → P ≠ 2 → P = 3 ∨ P ∈ ℤ ≥ 5
13 12 imp ⊢ P ∈ ℙ ∧ P ≠ 2 → P = 3 ∨ P ∈ ℤ ≥ 5
14 4 13 sylbi ⊢ P ∈ ℙ ∖ 2 → P = 3 ∨ P ∈ ℤ ≥ 5
15 fldiv4p1lem1div2 ⊢ P = 3 ∨ P ∈ ℤ ≥ 5 → P 4 + 1 ≤ P − 1 2
16 1 14 15 3syl ⊢ φ → P 4 + 1 ≤ P − 1 2
17 2 oveq1i ⊢ M + 1 = P 4 + 1
18 16 17 3 3brtr4g ⊢ φ → M + 1 ≤ H