Metamath Proof Explorer


Theorem ufli

Description: Property of a set that satisfies the ultrafilter lemma. (Contributed by Mario Carneiro, 26-Aug-2015)

Ref Expression
Assertion ufli ( ( 𝑋 ∈ UFL ∧ 𝐹 ∈ ( Fil ‘ 𝑋 ) ) → ∃ 𝑓 ∈ ( UFil ‘ 𝑋 ) 𝐹 ⊆ 𝑓 )

Proof

Step Hyp Ref Expression
1 isufl ⊢ ( 𝑋 ∈ UFL → ( 𝑋 ∈ UFL ↔ ∀ 𝑔 ∈ ( Fil ‘ 𝑋 ) ∃ 𝑓 ∈ ( UFil ‘ 𝑋 ) 𝑔 ⊆ 𝑓 ) )
2 1 ibi ⊢ ( 𝑋 ∈ UFL → ∀ 𝑔 ∈ ( Fil ‘ 𝑋 ) ∃ 𝑓 ∈ ( UFil ‘ 𝑋 ) 𝑔 ⊆ 𝑓 )
3 sseq1 ⊢ ( 𝑔 = 𝐹 → ( 𝑔 ⊆ 𝑓 ↔ 𝐹 ⊆ 𝑓 ) )
4 3 rexbidv ⊢ ( 𝑔 = 𝐹 → ( ∃ 𝑓 ∈ ( UFil ‘ 𝑋 ) 𝑔 ⊆ 𝑓 ↔ ∃ 𝑓 ∈ ( UFil ‘ 𝑋 ) 𝐹 ⊆ 𝑓 ) )
5 4 rspccva ⊢ ( ( ∀ 𝑔 ∈ ( Fil ‘ 𝑋 ) ∃ 𝑓 ∈ ( UFil ‘ 𝑋 ) 𝑔 ⊆ 𝑓 ∧ 𝐹 ∈ ( Fil ‘ 𝑋 ) ) → ∃ 𝑓 ∈ ( UFil ‘ 𝑋 ) 𝐹 ⊆ 𝑓 )
6 2 5 sylan ⊢ ( ( 𝑋 ∈ UFL ∧ 𝐹 ∈ ( Fil ‘ 𝑋 ) ) → ∃ 𝑓 ∈ ( UFil ‘ 𝑋 ) 𝐹 ⊆ 𝑓 )