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Proof step 4 from proof substring-offset-rec (previous step)

⊦ str = "", substring(str, m0 + 1, n) = substring(str.rstring, m0, n)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
str = char + str0 ⊦ str = "", substring(str, m0 + 1, n) = substring(str0, m0, n)

Rule argument

spec: string-data inst: name: elim seq: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0) subst: [str, char, str0] → [str, char, str0] flags:

Additional information

Used simplifier rules: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0); (elim) Used terms: str, char, str0