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

⊦ substring(str + str0, n0 + # str, n) = substring(str0, n0, n)

User applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ str0, n0, n. substring(str1 + str0, n0 + # str1, n) = substring(str0, n0, n) ⊦ substring((char + str1) + str0, n0 + #(char + str1), n) = substring(str0, n0, n)

Rule argument

indvar: str

Additional information

Used simplifier rules: string freely generated by "", +;;%append-base: ⊦ "" + str = str; from specification string-append%assoc: ⊦ (str1 + str2) + str3 = str1 + str2 + str3; from specification string-append%Length-empty: ⊦ # "" = 0; from specification string-less%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2