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Proof step 10 from proof violations-exists (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
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

Rule argument

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

Additional information

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