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Proof step 4 from proof violations-exists (previous step)

violations(a + x0) = 0, n + 1 < #(a + x0), (a + x0)[n + 1] < (a + x0)[n], ∀ n. violations(x0) = 0 ∧ x0[n + 1] < x0[n] ∧ n + 1 < # x0 → x0 = [] ⊦ a + x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
less#(a, x0) = 0, violations(x0) = 0, n < # x0, x0[n] < (a ' + x0)[n] ⊦ x0 = [], ∃ n0. x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0

Rule argument

Additional information

Used simplifier rules: violations-rec: ⊦ violations(a ' + x) = less#(a, x) + violations(x);% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ #(a ' + x) = (# x)+1; (s)% ⊦ (a ' + x)[n + 1] = x[n]; (s)% ⊦ a ' + x ≠ []; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)