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Proof step 9 from proof violations-exists (previous step)

less#(a, x0) = 0, violations(x0) = 0, x0[n] < (a ' + x0).tail[n - 1], n < # x0 ⊦ a ' + 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 10:
less#(a, x0) = 0, violations(x0) = 0, n < # x0, x0[n] < x0[n - 1] ⊦ n = 0, x0 = [], ∃ n0. x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0

Rule argument

Additional information

Used simplifier rules: ⊦ (a ' + x).tail = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)