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

∀ str0, n0, n. substring(str1 + str0, n0 + # str1, n) = substring(str0, n0, n) ⊦ (char ' + str1) + str0 = "", substring(((char ' + str1) + str0).rstring, n0 + # str1, n) = substring(str0, n0, n)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
(char ' + str1) + str0 = char0 + str2, ∀ str0, n0, n. substring(str1 + str0, n0 + # str1, n) = substring(str0, n0, n) ⊦ (char ' + str1) + str0 = "", substring(str2, n0 + # str1, n) = substring(str0, n0, n)

Rule argument

spec: string-data inst: name: elim seq: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0) subst: [str, char, str0] → [(char ' + str1) + str0, char0, str] flags:

Additional information

Used simplifier rules: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0); (elim) Used terms: (char ' + str1) + str0, char0, str