Metamath Proof Explorer


Theorem 2lgslem2

Description: Lemma 2 for 2lgs . (Contributed by AV, 20-Jun-2021)

Ref Expression
Hypothesis 2lgslem2.n ⊢ 𝑁 = ( ( ( 𝑃 − 1 ) / 2 ) − ( ⌊ ‘ ( 𝑃 / 4 ) ) )
Assertion 2lgslem2 ( ( 𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃 ) → 𝑁 ∈ ℤ )

Proof

Step Hyp Ref Expression
1 2lgslem2.n ⊢ 𝑁 = ( ( ( 𝑃 − 1 ) / 2 ) − ( ⌊ ‘ ( 𝑃 / 4 ) ) )
2 simpl ⊢ ( ( 𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃 ) → 𝑃 ∈ ℙ )
3 elsng ⊢ ( 𝑃 ∈ ℙ → ( 𝑃 ∈ { 2 } ↔ 𝑃 = 2 ) )
4 z2even ⊢ 2 ∥ 2
5 breq2 ⊢ ( 𝑃 = 2 → ( 2 ∥ 𝑃 ↔ 2 ∥ 2 ) )
6 4 5 mpbiri ⊢ ( 𝑃 = 2 → 2 ∥ 𝑃 )
7 3 6 biimtrdi ⊢ ( 𝑃 ∈ ℙ → ( 𝑃 ∈ { 2 } → 2 ∥ 𝑃 ) )
8 7 con3dimp ⊢ ( ( 𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃 ) → ¬ 𝑃 ∈ { 2 } )
9 2 8 eldifd ⊢ ( ( 𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃 ) → 𝑃 ∈ ( ℙ ∖ { 2 } ) )
10 oddprm ⊢ ( 𝑃 ∈ ( ℙ ∖ { 2 } ) → ( ( 𝑃 − 1 ) / 2 ) ∈ ℕ )
11 10 nnzd ⊢ ( 𝑃 ∈ ( ℙ ∖ { 2 } ) → ( ( 𝑃 − 1 ) / 2 ) ∈ ℤ )
12 9 11 syl ⊢ ( ( 𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃 ) → ( ( 𝑃 − 1 ) / 2 ) ∈ ℤ )
13 prmz ⊢ ( 𝑃 ∈ ℙ → 𝑃 ∈ ℤ )
14 13 zred ⊢ ( 𝑃 ∈ ℙ → 𝑃 ∈ ℝ )
15 4re ⊢ 4 ∈ ℝ
16 15 a1i ⊢ ( 𝑃 ∈ ℙ → 4 ∈ ℝ )
17 4ne0 ⊢ 4 ≠ 0
18 17 a1i ⊢ ( 𝑃 ∈ ℙ → 4 ≠ 0 )
19 14 16 18 redivcld ⊢ ( 𝑃 ∈ ℙ → ( 𝑃 / 4 ) ∈ ℝ )
20 19 flcld ⊢ ( 𝑃 ∈ ℙ → ( ⌊ ‘ ( 𝑃 / 4 ) ) ∈ ℤ )
21 20 adantr ⊢ ( ( 𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃 ) → ( ⌊ ‘ ( 𝑃 / 4 ) ) ∈ ℤ )
22 12 21 zsubcld ⊢ ( ( 𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃 ) → ( ( ( 𝑃 − 1 ) / 2 ) − ( ⌊ ‘ ( 𝑃 / 4 ) ) ) ∈ ℤ )
23 1 22 eqeltrid ⊢ ( ( 𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃 ) → 𝑁 ∈ ℤ )