Metamath Proof Explorer


Theorem filin

Description: A filter is closed under taking intersections. (Contributed by FL, 20-Jul-2007) (Revised by Stefan O'Rear, 28-Jul-2015)

Ref Expression
Assertion filin ⊢ F ∈ Fil ⁡ X ∧ A ∈ F ∧ B ∈ F → A ∩ B ∈ F

Proof

Step Hyp Ref Expression
1 filfbas ⊢ F ∈ Fil ⁡ X → F ∈ fBas ⁡ X
2 fbasssin ⊢ F ∈ fBas ⁡ X ∧ A ∈ F ∧ B ∈ F → ∃ x ∈ F x ⊆ A ∩ B
3 1 2 syl3an1 ⊢ F ∈ Fil ⁡ X ∧ A ∈ F ∧ B ∈ F → ∃ x ∈ F x ⊆ A ∩ B
4 inss1 ⊢ A ∩ B ⊆ A
5 filelss ⊢ F ∈ Fil ⁡ X ∧ A ∈ F → A ⊆ X
6 4 5 sstrid ⊢ F ∈ Fil ⁡ X ∧ A ∈ F → A ∩ B ⊆ X
7 filss ⊢ F ∈ Fil ⁡ X ∧ x ∈ F ∧ A ∩ B ⊆ X ∧ x ⊆ A ∩ B → A ∩ B ∈ F
8 7 3exp2 ⊢ F ∈ Fil ⁡ X → x ∈ F → A ∩ B ⊆ X → x ⊆ A ∩ B → A ∩ B ∈ F
9 8 com23 ⊢ F ∈ Fil ⁡ X → A ∩ B ⊆ X → x ∈ F → x ⊆ A ∩ B → A ∩ B ∈ F
10 9 imp ⊢ F ∈ Fil ⁡ X ∧ A ∩ B ⊆ X → x ∈ F → x ⊆ A ∩ B → A ∩ B ∈ F
11 10 rexlimdv ⊢ F ∈ Fil ⁡ X ∧ A ∩ B ⊆ X → ∃ x ∈ F x ⊆ A ∩ B → A ∩ B ∈ F
12 6 11 syldan ⊢ F ∈ Fil ⁡ X ∧ A ∈ F → ∃ x ∈ F x ⊆ A ∩ B → A ∩ B ∈ F
13 12 3adant3 ⊢ F ∈ Fil ⁡ X ∧ A ∈ F ∧ B ∈ F → ∃ x ∈ F x ⊆ A ∩ B → A ∩ B ∈ F
14 3 13 mpd ⊢ F ∈ Fil ⁡ X ∧ A ∈ F ∧ B ∈ F → A ∩ B ∈ F