Metamath Proof Explorer

Theorem pfxfn

Description: Value of the prefix extractor as function with domain. (Contributed by AV, 2-May-2020)

Ref Expression
Assertion pfxfn 
|- ( ( S e. Word V /\ L e. ( 0 ... ( # ` S ) ) ) -> ( S prefix L ) Fn ( 0 ..^ L ) )

Proof

Step Hyp Ref Expression
1 pfxf 
 |-  ( ( S e. Word V /\ L e. ( 0 ... ( # ` S ) ) ) -> ( S prefix L ) : ( 0 ..^ L ) --> V )
2 1 ffnd 
 |-  ( ( S e. Word V /\ L e. ( 0 ... ( # ` S ) ) ) -> ( S prefix L ) Fn ( 0 ..^ L ) )