Metamath Proof Explorer


Theorem suprcld

Description: Natural deduction form of suprcl . (Contributed by Stanislas Polu, 9-Mar-2020)

Ref Expression
Hypotheses suprcld.2 ⊢ φ → A ⊆ ℝ
suprcld.1 ⊢ φ → A ≠ ∅
suprcld.4 ⊢ φ → ∃ x ∈ ℝ ∀ y ∈ A y ≤ x
Assertion suprcld ⊢ φ → sup A ℝ < ∈ ℝ

Proof

Step Hyp Ref Expression
1 suprcld.2 ⊢ φ → A ⊆ ℝ
2 suprcld.1 ⊢ φ → A ≠ ∅
3 suprcld.4 ⊢ φ → ∃ x ∈ ℝ ∀ y ∈ A y ≤ x
4 suprcl ⊢ A ⊆ ℝ ∧ A ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ A y ≤ x → sup A ℝ < ∈ ℝ
5 1 2 3 4 syl3anc ⊢ φ → sup A ℝ < ∈ ℝ