Metamath Proof Explorer


Theorem 4sqlem7

Description: Lemma for 4sq . (Contributed by Mario Carneiro, 15-Jul-2014)

Ref Expression
Hypotheses 4sqlem5.2 ⊢ φ → A ∈ ℤ
4sqlem5.3 ⊢ φ → M ∈ ℕ
4sqlem5.4 ⊢ B = A + M 2 mod M − M 2
Assertion 4sqlem7 ⊢ φ → B 2 ≤ M 2 2 2

Proof

Step Hyp Ref Expression
1 4sqlem5.2 ⊢ φ → A ∈ ℤ
2 4sqlem5.3 ⊢ φ → M ∈ ℕ
3 4sqlem5.4 ⊢ B = A + M 2 mod M − M 2
4 1 2 3 4sqlem5 ⊢ φ → B ∈ ℤ ∧ A − B M ∈ ℤ
5 4 simpld ⊢ φ → B ∈ ℤ
6 5 zred ⊢ φ → B ∈ ℝ
7 2 nnrpd ⊢ φ → M ∈ ℝ +
8 7 rphalfcld ⊢ φ → M 2 ∈ ℝ +
9 8 rpred ⊢ φ → M 2 ∈ ℝ
10 1 2 3 4sqlem6 ⊢ φ → − M 2 ≤ B ∧ B < M 2
11 10 simprd ⊢ φ → B < M 2
12 6 9 11 ltled ⊢ φ → B ≤ M 2
13 10 simpld ⊢ φ → − M 2 ≤ B
14 9 6 13 lenegcon1d ⊢ φ → − B ≤ M 2
15 8 rpge0d ⊢ φ → 0 ≤ M 2
16 lenegsq ⊢ B ∈ ℝ ∧ M 2 ∈ ℝ ∧ 0 ≤ M 2 → B ≤ M 2 ∧ − B ≤ M 2 ↔ B 2 ≤ M 2 2
17 6 9 15 16 syl3anc ⊢ φ → B ≤ M 2 ∧ − B ≤ M 2 ↔ B 2 ≤ M 2 2
18 12 14 17 mpbi2and ⊢ φ → B 2 ≤ M 2 2
19 2cnd ⊢ φ → 2 ∈ ℂ
20 19 sqvald ⊢ φ → 2 2 = 2 ⋅ 2
21 20 oveq2d ⊢ φ → M 2 2 2 = M 2 2 ⋅ 2
22 2 nncnd ⊢ φ → M ∈ ℂ
23 2ne0 ⊢ 2 ≠ 0
24 23 a1i ⊢ φ → 2 ≠ 0
25 22 19 24 sqdivd ⊢ φ → M 2 2 = M 2 2 2
26 22 sqcld ⊢ φ → M 2 ∈ ℂ
27 26 19 19 24 24 divdiv1d ⊢ φ → M 2 2 2 = M 2 2 ⋅ 2
28 21 25 27 3eqtr4d ⊢ φ → M 2 2 = M 2 2 2
29 18 28 breqtrd ⊢ φ → B 2 ≤ M 2 2 2