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

m = m0 + 1 ⊦ str = "", m = 0, substring(str, m, n) = substring(str.rstring, m0, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ str = "", substring(str, m0 + 1, n) = substring(str.rstring, m0, 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%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2