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Proof step 8 from proof violations-exists (previous step)

less#(a, x0) = 0, violations(x0) = 0, x0.head < a ⊦ x0 = [], ∃ n0. x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [x0, a0, y]

Additional information

Used simplifier rules: violations-rec: ⊦ violations(a ' + x) = less#(a, x) + violations(x);%less-rec: ⊦ less#(a, b ' + x) = (b < a ⊃ 1;0) + less#(a, x);% ⊦ a + x = a ' + x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x0, a0, y