Metamath Proof Explorer


Theorem sn-suprcld

Description: suprcld without ax-mulcom , proven trivially from sn-sup3d . (Contributed by SN, 29-Jun-2025)

Ref Expression
Hypotheses sn-sup3d.1 ⊢ φ → A ⊆ ℝ
sn-sup3d.2 ⊢ φ → A ≠ ∅
sn-sup3d.3 ⊢ φ → ∃ x ∈ ℝ ∀ y ∈ A y ≤ x
Assertion sn-suprcld ⊢ φ → sup A ℝ < ∈ ℝ

Proof

Step Hyp Ref Expression
1 sn-sup3d.1 ⊢ φ → A ⊆ ℝ
2 sn-sup3d.2 ⊢ φ → A ≠ ∅
3 sn-sup3d.3 ⊢ φ → ∃ x ∈ ℝ ∀ y ∈ A y ≤ x
4 ltso ⊢ < Or ℝ
5 4 a1i ⊢ φ → < Or ℝ
6 1 2 3 sn-sup3d ⊢ φ → ∃ x ∈ ℝ ∀ y ∈ A ¬ x < y ∧ ∀ y ∈ ℝ y < x → ∃ z ∈ A y < z
7 5 6 supcl ⊢ φ → sup A ℝ < ∈ ℝ