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Proof step 9 from proof substring-zero-rec (previous step)

⊦ char ' + str0 = "", l→s(sublist(0, m + 1, char ' + s→l(str0))) = char ' + substring(str0, 0, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ chars ≠ [] → sublist(0, n + 1, chars) = chars.first + sublist(0, n, chars.rest)
Proof step 11:
char ' + s→l(str0) ≠ [] ⊦ char ' + str0 = "", l→s((char ' + s→l(str0)).first + sublist(0, m, (char ' + s→l(str0)).rest)) = char ' + substring(str0, 0, m)

Rule argument

spec: list-del inst: charlist name: sub-zerorec seq: ⊦ chars ≠ [] → sublist(0, n + 1, chars) = chars.first + sublist(0, n, chars.rest) subst: [chars, n] → [char ' + s→l(str0), m] flags: rewrite

Additional information

Used simplifier rules: app-02-ant: ⊦ ¬ chars + chars0 ≠ [] ↔ chars = [] ∧ chars0 = []; from specification list[charlist]%e: ⊦ char ' ≠ []; from specification list[charlist]%a: ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1; from specification list[charlist] Used terms: char ' + s→l(str0), m