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Proof step 3 from proof violations-swap (previous step)

x[n + 1] < x[n], n + 1 < # x ⊦ violations(x[n, x[n + 1]][n + 1, x[n]]) = violations(x) - 1, x = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
violations(x) = m + 1, n + 1 < # x, x[n + 1] < x[n] ⊦ violations(x[n, x[n + 1]][n + 1, x[n]]) = m, violations(x) = 0, x = []

Rule argument

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

Additional information

Used simplifier rules: violations-exists: n + 1 < # x ⊦ x[n + 1] < x[n] → violations(x) ≠ 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: violations(x), m