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Proof step 3 from proof substring-app-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ substring(str + str0, n0 + # str, n) = substring(str0, n0, n)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat