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 e. ( Fil ` X ) /\ A e. F ) -> ( F |`t A ) C_ F )

Proof

Step Hyp Ref Expression
1 restval
 |-  ( ( F e. ( Fil ` X ) /\ A e. F ) -> ( F |`t A ) = ran ( x e. F |-> ( x i^i A ) ) )
2 filin
 |-  ( ( F e. ( Fil ` X ) /\ x e. F /\ A e. F ) -> ( x i^i A ) e. F )
3 2 3expa
 |-  ( ( ( F e. ( Fil ` X ) /\ x e. F ) /\ A e. F ) -> ( x i^i A ) e. F )
4 3 an32s
 |-  ( ( ( F e. ( Fil ` X ) /\ A e. F ) /\ x e. F ) -> ( x i^i A ) e. F )
5 4 fmpttd
 |-  ( ( F e. ( Fil ` X ) /\ A e. F ) -> ( x e. F |-> ( x i^i A ) ) : F --> F )
6 5 frnd
 |-  ( ( F e. ( Fil ` X ) /\ A e. F ) -> ran ( x e. F |-> ( x i^i A ) ) C_ F )
7 1 6 eqsstrd
 |-  ( ( F e. ( Fil ` X ) /\ A e. F ) -> ( F |`t A ) C_ F )