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Proof step 2 from proof violations-exists (previous step)

violations(x) = 0, x[n + 1] < x[n], n + 1 < # x ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
violations(x) = 0, x[n + 1] < x[n], n + 1 < # x ⊦ x = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic