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Proof step 15 from proof violations-swap (previous step)

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 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
less#(a, x0) + violations(x0) = m + 1, n = m1 + 1, x0[n] < x0[m1], 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 ⊦ n = 0, violations((a ' + x0)[n, x0[n]][n + 1, x0[m1]]) = m, x0 = []

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m1]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m1