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Proof step 2 from proof substring-offset-rec (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
m = m0 + 1 ⊦ str = "", m = 0, substring(str, m, n) = substring(str.rstring, m0, n)

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m0] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0