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Proof step 14 from proof violations-swap (previous step)

less#(a, x0) + violations(x0) = m + 1, x0[n] < (a ' + x0).tail[n - 1], n < # x0, ∀ m0, n0. violations(x0) = m0 + 1 ∧ x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0 → violations(x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = m0 ⊦ a ' + x0 = [], n = 0, violations((a ' + x0)[n, x0[n]][n + 1, (a ' + x0).tail[n - 1]]) = m, x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
less#(a, x0) + violations(x0) = m + 1, n < # x0, x0[n] < x0[n - 1], ∀ m0, n0. violations(x0) = m0 + 1 ∧ x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0 → violations(x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = m0 ⊦ n = 0, violations((a ' + x0)[n, x0[n]][n + 1, x0[n - 1]]) = m, x0 = []

Rule argument

Additional information

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