Theorem swrdvalfn

Description: Value of the subword extractor as function with domain. (Contributed by Alexander van der Vekens, 28-Mar-2018) (Proof shortened by AV, 2-May-2020)

	Ref	Expression
Assertion	swrdvalfn	⊢ ( ( 𝑆 ∈ Word 𝑉 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 substr ⟨ 𝐹 , 𝐿 ⟩ ) Fn ( 0 ..^ ( 𝐿 − 𝐹 ) ) )

Proof

Step	Hyp	Ref	Expression
1		swrdf	⊢ ( ( 𝑆 ∈ Word 𝑉 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 substr ⟨ 𝐹 , 𝐿 ⟩ ) : ( 0 ..^ ( 𝐿 − 𝐹 ) ) ⟶ 𝑉 )
2	1	ffnd	⊢ ( ( 𝑆 ∈ Word 𝑉 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 substr ⟨ 𝐹 , 𝐿 ⟩ ) Fn ( 0 ..^ ( 𝐿 − 𝐹 ) ) )