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 ≠ ∅ → x ∈ F

Proof

Step Hyp Ref Expression
1 df-fil ⊢ Fil = z ∈ V ⟼ f ∈ fBas ⁡ z | ∀ x ∈ 𝒫 z f ∩ 𝒫 x ≠ ∅ → x ∈ f
2 pweq ⊢ z = X → 𝒫 z = 𝒫 X
3 2 adantr ⊢ z = X ∧ f = 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 → x ∈ f ↔ x ∈ F
7 5 6 imbi12d ⊢ f = F → f ∩ 𝒫 x ≠ ∅ → x ∈ f ↔ F ∩ 𝒫 x ≠ ∅ → x ∈ F
8 7 adantl ⊢ z = X ∧ f = F → f ∩ 𝒫 x ≠ ∅ → x ∈ f ↔ F ∩ 𝒫 x ≠ ∅ → x ∈ F
9 3 8 raleqbidv ⊢ z = X ∧ f = F → ∀ x ∈ 𝒫 z f ∩ 𝒫 x ≠ ∅ → x ∈ f ↔ ∀ x ∈ 𝒫 X F ∩ 𝒫 x ≠ ∅ → x ∈ F
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 ≠ ∅ → x ∈ F