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 ∈ Word 𝑉𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 prefix 𝐿 ) Fn ( 0 ..^ 𝐿 ) )

Proof

Step Hyp Ref Expression
1 pfxf ( ( 𝑆 ∈ Word 𝑉𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 prefix 𝐿 ) : ( 0 ..^ 𝐿 ) ⟶ 𝑉 )
2 1 ffnd ( ( 𝑆 ∈ Word 𝑉𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 prefix 𝐿 ) Fn ( 0 ..^ 𝐿 ) )