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Proof step 4 from proof substring-all (previous step)

⊦ l→s(sublist(0, # str, s→l(str))) = str

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ # chars = n → sublist(0, n, chars) = chars

Rule argument

spec: list-del inst: charlist name: sublist-all seq: ⊦ # chars = n → sublist(0, n, chars) = chars subst: [chars, n] → [s→l(str), # str] flags: rewrite

Additional information

Used simplifier rules: s2l-length: ⊦ # s→l(str) = # str; from specification string2list%string2string: ⊦ l→s(s→l(str)) = str; from specification string2list Used terms: s→l(str), # str