Metamath Proof Explorer


Theorem 01sqrexlem3

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 01sqrexlem3 ⊢ A ∈ ℝ + ∧ A ≤ 1 → S ⊆ ℝ ∧ S ≠ ∅ ∧ ∃ z ∈ ℝ ∀ y ∈ S y ≤ z

Proof

Step Hyp Ref Expression
1 01sqrexlem1.1 ⊢ S = x ∈ ℝ + | x 2 ≤ A
2 01sqrexlem1.2 ⊢ B = sup S ℝ <
3 ssrab2 ⊢ x ∈ ℝ + | x 2 ≤ A ⊆ ℝ +
4 rpssre ⊢ ℝ + ⊆ ℝ
5 3 4 sstri ⊢ x ∈ ℝ + | x 2 ≤ A ⊆ ℝ
6 1 5 eqsstri ⊢ S ⊆ ℝ
7 6 a1i ⊢ A ∈ ℝ + ∧ A ≤ 1 → S ⊆ ℝ
8 1 2 01sqrexlem2 ⊢ A ∈ ℝ + ∧ A ≤ 1 → A ∈ S
9 8 ne0d ⊢ A ∈ ℝ + ∧ A ≤ 1 → S ≠ ∅
10 1re ⊢ 1 ∈ ℝ
11 1 2 01sqrexlem1 ⊢ A ∈ ℝ + ∧ A ≤ 1 → ∀ y ∈ S y ≤ 1
12 brralrspcev ⊢ 1 ∈ ℝ ∧ ∀ y ∈ S y ≤ 1 → ∃ z ∈ ℝ ∀ y ∈ S y ≤ z
13 10 11 12 sylancr ⊢ A ∈ ℝ + ∧ A ≤ 1 → ∃ z ∈ ℝ ∀ y ∈ S y ≤ z
14 7 9 13 3jca ⊢ A ∈ ℝ + ∧ A ≤ 1 → S ⊆ ℝ ∧ S ≠ ∅ ∧ ∃ z ∈ ℝ ∀ y ∈ S y ≤ z