Metamath Proof Explorer


Theorem trfilss

Description: If A is a member of the filter, then the filter truncated to A is a subset of the original filter. (Contributed by Mario Carneiro, 15-Oct-2015)

Ref Expression
Assertion trfilss ⊢ F ∈ Fil ⁡ X ∧ A ∈ F → F ↾ 𝑡 A ⊆ F

Proof

Step Hyp Ref Expression
1 restval ⊢ F ∈ Fil ⁡ X ∧ A ∈ F → F ↾ 𝑡 A = ran ⁡ x ∈ F ⟼ x ∩ A
2 filin ⊢ F ∈ Fil ⁡ X ∧ x ∈ F ∧ A ∈ F → x ∩ A ∈ F
3 2 3expa ⊢ F ∈ Fil ⁡ X ∧ x ∈ F ∧ A ∈ F → x ∩ A ∈ F
4 3 an32s ⊢ F ∈ Fil ⁡ X ∧ A ∈ F ∧ x ∈ F → x ∩ A ∈ F
5 4 fmpttd ⊢ F ∈ Fil ⁡ X ∧ A ∈ F → x ∈ F ⟼ x ∩ A : F ⟶ F
6 5 frnd ⊢ F ∈ Fil ⁡ X ∧ A ∈ F → ran ⁡ x ∈ F ⟼ x ∩ A ⊆ F
7 1 6 eqsstrd ⊢ F ∈ Fil ⁡ X ∧ A ∈ F → F ↾ 𝑡 A ⊆ F