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Proof step 2 from proof butlast-length (previous step)

⊦ str = "", # str.butlast = # str - 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
# str = m + 1 ∧ # str ≠ 0 ⊦ str = "", # str.butlast = m

Rule argument

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

Additional information

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