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Proof step 17 from proof violations-swap (previous step)

less#(a, x0) + violations(x0) = m + 1, m1 + 1 < # x0, x0[m1 + 1] < x0[m1], ∀ 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 ⊦ violations((a ' + x0[m1, x0[m1 + 1]])[2 + m1, x0[m1]]) = m, x0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 18:
⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.head ' + x.tail[n -1, a]
Proof step 19:
less#(a, x0) + violations(x0) = m + 1, x0[m1 + 1] < x0[m1], m1 + 1 < # 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 ⊦ a ' + x0[m1, x0[m1 + 1]] = [], 2 + m1 = 0, violations((a ' + x0[m1, x0[m1 + 1]]).head ' + (a ' + x0[m1, x0[m1 + 1]]).tail[(2 + m1)-1, x0[m1]]) = m, x0 = []

Rule argument

spec: list-del name: put-rec seq: ⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.head ' + x.tail[n -1, a] subst: [a, n, x] → [x0[m1], 2 + m1, a ' + x0[m1, x0[m1 + 1]]] flags: <all paths>

Additional information

Used simplifier rules: n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0[m1], 2 + m1, a ' + x0[m1, x0[m1 + 1]]