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

⊦ char ' + str0 = "", l→s(sublist(m0 + 1, n, char ' + s→l(str0))) = substring(str0, m0, n)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ sublist(m + 1, n, chars) = sublist(m, n, chars.rest)
Proof step 11:
⊦ char ' + str0 = "", l→s(sublist(m0, n, (char ' + s→l(str0)).rest)) = substring(str0, m0, n)

Rule argument

spec: list-del inst: charlist name: sub-rec seq: ⊦ sublist(m + 1, n, chars) = sublist(m, n, chars.rest) subst: [chars, n, m] → [char ' + s→l(str0), n, m0] flags: rewrite

Additional information

Used terms: char ' + s→l(str0), n, m0