Metamath Proof Explorer


Theorem 01sqrexlem2

Description: Lemma for 01sqrex . (Contributed by Mario Carneiro, 10-Jul-2013)

Ref Expression
Hypotheses 01sqrexlem1.1 ⊢ S = x ∈ ℝ + | x 2 ≤ A
01sqrexlem1.2 ⊢ B = sup S ℝ <
Assertion 01sqrexlem2 ⊢ A ∈ ℝ + ∧ A ≤ 1 → A ∈ S

Proof

Step Hyp Ref Expression
1 01sqrexlem1.1 ⊢ S = x ∈ ℝ + | x 2 ≤ A
2 01sqrexlem1.2 ⊢ B = sup S ℝ <
3 simpl ⊢ A ∈ ℝ + ∧ A ≤ 1 → A ∈ ℝ +
4 rpre ⊢ A ∈ ℝ + → A ∈ ℝ
5 rpgt0 ⊢ A ∈ ℝ + → 0 < A
6 1re ⊢ 1 ∈ ℝ
7 lemul1 ⊢ A ∈ ℝ ∧ 1 ∈ ℝ ∧ A ∈ ℝ ∧ 0 < A → A ≤ 1 ↔ A ⁢ A ≤ 1 ⁢ A
8 6 7 mp3an2 ⊢ A ∈ ℝ ∧ A ∈ ℝ ∧ 0 < A → A ≤ 1 ↔ A ⁢ A ≤ 1 ⁢ A
9 4 4 5 8 syl12anc ⊢ A ∈ ℝ + → A ≤ 1 ↔ A ⁢ A ≤ 1 ⁢ A
10 9 biimpa ⊢ A ∈ ℝ + ∧ A ≤ 1 → A ⁢ A ≤ 1 ⁢ A
11 rpcn ⊢ A ∈ ℝ + → A ∈ ℂ
12 11 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 → A ∈ ℂ
13 sqval ⊢ A ∈ ℂ → A 2 = A ⁢ A
14 13 eqcomd ⊢ A ∈ ℂ → A ⁢ A = A 2
15 12 14 syl ⊢ A ∈ ℝ + ∧ A ≤ 1 → A ⁢ A = A 2
16 11 mullidd ⊢ A ∈ ℝ + → 1 ⁢ A = A
17 16 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 → 1 ⁢ A = A
18 10 15 17 3brtr3d ⊢ A ∈ ℝ + ∧ A ≤ 1 → A 2 ≤ A
19 oveq1 ⊢ x = A → x 2 = A 2
20 19 breq1d ⊢ x = A → x 2 ≤ A ↔ A 2 ≤ A
21 20 1 elrab2 ⊢ A ∈ S ↔ A ∈ ℝ + ∧ A 2 ≤ A
22 3 18 21 sylanbrc ⊢ A ∈ ℝ + ∧ A ≤ 1 → A ∈ S