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Proof step 11 from proof violations-exists (previous step)

less#(a, x0) = 0, n = m + 1, violations(x0) = 0, x0[n] < x0[m], n < # x0 ⊦ n = 0, x0 = [], ∃ n0. x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
less#(a, x0) = 0, violations(x0) = 0, m + 1 < # x0, x0[m + 1] < x0[m] ⊦ x0 = [], ∃ n0. x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)