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 )