Metamath Proof Explorer


Theorem dchrisum0lem1a

Description: Lemma for dchrisum0lem1 . (Contributed by Mario Carneiro, 7-Jun-2016)

Ref Expression
Assertion dchrisum0lem1a ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X ≤ X 2 D ∧ X 2 D ∈ ℤ ≥ X

Proof

Step Hyp Ref Expression
1 elfznn ⊢ D ∈ 1 … X → D ∈ ℕ
2 1 adantl ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → D ∈ ℕ
3 2 nnred ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → D ∈ ℝ
4 simpr ⊢ φ ∧ X ∈ ℝ + → X ∈ ℝ +
5 4 rpregt0d ⊢ φ ∧ X ∈ ℝ + → X ∈ ℝ ∧ 0 < X
6 5 adantr ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X ∈ ℝ ∧ 0 < X
7 6 simpld ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X ∈ ℝ
8 4 adantr ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X ∈ ℝ +
9 8 rpge0d ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → 0 ≤ X
10 4 rpred ⊢ φ ∧ X ∈ ℝ + → X ∈ ℝ
11 fznnfl ⊢ X ∈ ℝ → D ∈ 1 … X ↔ D ∈ ℕ ∧ D ≤ X
12 10 11 syl ⊢ φ ∧ X ∈ ℝ + → D ∈ 1 … X ↔ D ∈ ℕ ∧ D ≤ X
13 12 simplbda ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → D ≤ X
14 3 7 7 9 13 lemul2ad ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X ⁢ D ≤ X ⁢ X
15 rpcn ⊢ X ∈ ℝ + → X ∈ ℂ
16 15 adantl ⊢ φ ∧ X ∈ ℝ + → X ∈ ℂ
17 16 sqvald ⊢ φ ∧ X ∈ ℝ + → X 2 = X ⁢ X
18 17 adantr ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X 2 = X ⁢ X
19 14 18 breqtrrd ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X ⁢ D ≤ X 2
20 2z ⊢ 2 ∈ ℤ
21 rpexpcl ⊢ X ∈ ℝ + ∧ 2 ∈ ℤ → X 2 ∈ ℝ +
22 4 20 21 sylancl ⊢ φ ∧ X ∈ ℝ + → X 2 ∈ ℝ +
23 22 rpred ⊢ φ ∧ X ∈ ℝ + → X 2 ∈ ℝ
24 23 adantr ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X 2 ∈ ℝ
25 2 nnrpd ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → D ∈ ℝ +
26 7 24 25 lemuldivd ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X ⁢ D ≤ X 2 ↔ X ≤ X 2 D
27 19 26 mpbid ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X ≤ X 2 D
28 nndivre ⊢ X 2 ∈ ℝ ∧ D ∈ ℕ → X 2 D ∈ ℝ
29 23 1 28 syl2an ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X 2 D ∈ ℝ
30 flword2 ⊢ X ∈ ℝ ∧ X 2 D ∈ ℝ ∧ X ≤ X 2 D → X 2 D ∈ ℤ ≥ X
31 7 29 27 30 syl3anc ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X 2 D ∈ ℤ ≥ X
32 27 31 jca ⊢ φ ∧ X ∈ ℝ + ∧ D ∈ 1 … X → X ≤ X 2 D ∧ X 2 D ∈ ℤ ≥ X