Metamath Proof Explorer


Theorem isfil

Description: The predicate "is a filter." (Contributed by FL, 20-Jul-2007) (Revised by Mario Carneiro, 28-Jul-2015)

Ref Expression
Assertion isfil ( 𝐹 ∈ ( Fil ‘ 𝑋 ) ↔ ( 𝐹 ∈ ( fBas ‘ 𝑋 ) ∧ ∀ 𝑥 ∈ 𝒫 𝑋 ( ( 𝐹 ∩ 𝒫 𝑥 ) ≠ ∅ → 𝑥𝐹 ) ) )

Proof

Step Hyp Ref Expression
1 df-fil Fil = ( 𝑧 ∈ V ↦ { 𝑓 ∈ ( fBas ‘ 𝑧 ) ∣ ∀ 𝑥 ∈ 𝒫 𝑧 ( ( 𝑓 ∩ 𝒫 𝑥 ) ≠ ∅ → 𝑥𝑓 ) } )
2 pweq ( 𝑧 = 𝑋 → 𝒫 𝑧 = 𝒫 𝑋 )
3 2 adantr ( ( 𝑧 = 𝑋𝑓 = 𝐹 ) → 𝒫 𝑧 = 𝒫 𝑋 )
4 ineq1 ( 𝑓 = 𝐹 → ( 𝑓 ∩ 𝒫 𝑥 ) = ( 𝐹 ∩ 𝒫 𝑥 ) )
5 4 neeq1d ( 𝑓 = 𝐹 → ( ( 𝑓 ∩ 𝒫 𝑥 ) ≠ ∅ ↔ ( 𝐹 ∩ 𝒫 𝑥 ) ≠ ∅ ) )
6 eleq2 ( 𝑓 = 𝐹 → ( 𝑥𝑓𝑥𝐹 ) )
7 5 6 imbi12d ( 𝑓 = 𝐹 → ( ( ( 𝑓 ∩ 𝒫 𝑥 ) ≠ ∅ → 𝑥𝑓 ) ↔ ( ( 𝐹 ∩ 𝒫 𝑥 ) ≠ ∅ → 𝑥𝐹 ) ) )
8 7 adantl ( ( 𝑧 = 𝑋𝑓 = 𝐹 ) → ( ( ( 𝑓 ∩ 𝒫 𝑥 ) ≠ ∅ → 𝑥𝑓 ) ↔ ( ( 𝐹 ∩ 𝒫 𝑥 ) ≠ ∅ → 𝑥𝐹 ) ) )
9 3 8 raleqbidv ( ( 𝑧 = 𝑋𝑓 = 𝐹 ) → ( ∀ 𝑥 ∈ 𝒫 𝑧 ( ( 𝑓 ∩ 𝒫 𝑥 ) ≠ ∅ → 𝑥𝑓 ) ↔ ∀ 𝑥 ∈ 𝒫 𝑋 ( ( 𝐹 ∩ 𝒫 𝑥 ) ≠ ∅ → 𝑥𝐹 ) ) )
10 fveq2 ( 𝑧 = 𝑋 → ( fBas ‘ 𝑧 ) = ( fBas ‘ 𝑋 ) )
11 fvex ( fBas ‘ 𝑧 ) ∈ V
12 elfvdm ( 𝐹 ∈ ( fBas ‘ 𝑋 ) → 𝑋 ∈ dom fBas )
13 1 9 10 11 12 elmptrab2 ( 𝐹 ∈ ( Fil ‘ 𝑋 ) ↔ ( 𝐹 ∈ ( fBas ‘ 𝑋 ) ∧ ∀ 𝑥 ∈ 𝒫 𝑋 ( ( 𝐹 ∩ 𝒫 𝑥 ) ≠ ∅ → 𝑥𝐹 ) ) )