Metamath Proof Explorer


Theorem suprubd

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

Ref Expression
Hypotheses suprubd.1 ⊢ ( 𝜑 → 𝐴 ⊆ ℝ )
suprubd.2 ⊢ ( 𝜑 → 𝐴 ≠ ∅ )
suprubd.3 ⊢ ( 𝜑 → ∃ 𝑥 ∈ ℝ ∀ 𝑦 ∈ 𝐴 𝑦 ≤ 𝑥 )
suprubd.4 ⊢ ( 𝜑 → 𝐵 ∈ 𝐴 )
Assertion suprubd ( 𝜑 → 𝐵 ≤ sup ( 𝐴 , ℝ , < ) )

Proof

Step Hyp Ref Expression
1 suprubd.1 ⊢ ( 𝜑 → 𝐴 ⊆ ℝ )
2 suprubd.2 ⊢ ( 𝜑 → 𝐴 ≠ ∅ )
3 suprubd.3 ⊢ ( 𝜑 → ∃ 𝑥 ∈ ℝ ∀ 𝑦 ∈ 𝐴 𝑦 ≤ 𝑥 )
4 suprubd.4 ⊢ ( 𝜑 → 𝐵 ∈ 𝐴 )
5 suprub ⊢ ( ( ( 𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃ 𝑥 ∈ ℝ ∀ 𝑦 ∈ 𝐴 𝑦 ≤ 𝑥 ) ∧ 𝐵 ∈ 𝐴 ) → 𝐵 ≤ sup ( 𝐴 , ℝ , < ) )
6 1 2 3 4 5 syl31anc ⊢ ( 𝜑 → 𝐵 ≤ sup ( 𝐴 , ℝ , < ) )