Theorem pfxlen

Description: Length of a prefix. (Contributed by Stefan O'Rear, 24-Aug-2015) (Revised by AV, 2-May-2020)

	Ref	Expression
Assertion	pfxlen	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( ♯ ‘ ( 𝑆 prefix 𝐿 ) ) = 𝐿 )

Proof

Step	Hyp	Ref	Expression
1		pfxfn	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 prefix 𝐿 ) Fn ( 0 ..^ 𝐿 ) )
2		hashfn	⊢ ( ( 𝑆 prefix 𝐿 ) Fn ( 0 ..^ 𝐿 ) → ( ♯ ‘ ( 𝑆 prefix 𝐿 ) ) = ( ♯ ‘ ( 0 ..^ 𝐿 ) ) )
3	1 2	syl	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( ♯ ‘ ( 𝑆 prefix 𝐿 ) ) = ( ♯ ‘ ( 0 ..^ 𝐿 ) ) )
4		elfznn0	⊢ ( 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) → 𝐿 ∈ ℕ0 )
5	4	adantl	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → 𝐿 ∈ ℕ0 )
6		hashfzo0	⊢ ( 𝐿 ∈ ℕ0 → ( ♯ ‘ ( 0 ..^ 𝐿 ) ) = 𝐿 )
7	5 6	syl	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( ♯ ‘ ( 0 ..^ 𝐿 ) ) = 𝐿 )
8	3 7	eqtrd	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( ♯ ‘ ( 𝑆 prefix 𝐿 ) ) = 𝐿 )