Metamath Proof Explorer


Theorem glbelss

Description: A member of the domain of the greatest lower bound function is a subset of the base set. (Contributed by NM, 7-Sep-2018)

Ref Expression
Hypotheses glbs.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
glbs.l ⊢ ≤ = ( le ‘ 𝐾 )
glbs.g ⊢ 𝐺 = ( glb ‘ 𝐾 )
glbs.k ⊢ ( 𝜑 → 𝐾 ∈ 𝑉 )
glbs.s ⊢ ( 𝜑 → 𝑆 ∈ dom 𝐺 )
Assertion glbelss ( 𝜑 → 𝑆 ⊆ 𝐵 )

Proof

Step Hyp Ref Expression
1 glbs.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 glbs.l ⊢ ≤ = ( le ‘ 𝐾 )
3 glbs.g ⊢ 𝐺 = ( glb ‘ 𝐾 )
4 glbs.k ⊢ ( 𝜑 → 𝐾 ∈ 𝑉 )
5 glbs.s ⊢ ( 𝜑 → 𝑆 ∈ dom 𝐺 )
6 biid ⊢ ( ( ∀ 𝑦 ∈ 𝑆 𝑥 ≤ 𝑦 ∧ ∀ 𝑧 ∈ 𝐵 ( ∀ 𝑦 ∈ 𝑆 𝑧 ≤ 𝑦 → 𝑧 ≤ 𝑥 ) ) ↔ ( ∀ 𝑦 ∈ 𝑆 𝑥 ≤ 𝑦 ∧ ∀ 𝑧 ∈ 𝐵 ( ∀ 𝑦 ∈ 𝑆 𝑧 ≤ 𝑦 → 𝑧 ≤ 𝑥 ) ) )
7 1 2 3 6 4 glbeldm ⊢ ( 𝜑 → ( 𝑆 ∈ dom 𝐺 ↔ ( 𝑆 ⊆ 𝐵 ∧ ∃! 𝑥 ∈ 𝐵 ( ∀ 𝑦 ∈ 𝑆 𝑥 ≤ 𝑦 ∧ ∀ 𝑧 ∈ 𝐵 ( ∀ 𝑦 ∈ 𝑆 𝑧 ≤ 𝑦 → 𝑧 ≤ 𝑥 ) ) ) ) )
8 5 7 mpbid ⊢ ( 𝜑 → ( 𝑆 ⊆ 𝐵 ∧ ∃! 𝑥 ∈ 𝐵 ( ∀ 𝑦 ∈ 𝑆 𝑥 ≤ 𝑦 ∧ ∀ 𝑧 ∈ 𝐵 ( ∀ 𝑦 ∈ 𝑆 𝑧 ≤ 𝑦 → 𝑧 ≤ 𝑥 ) ) ) )
9 8 simpld ⊢ ( 𝜑 → 𝑆 ⊆ 𝐵 )