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Proof step 15 from proof sorted-next-1-swap (previous step)

n % 2 = 1, 2 * n0 + 1 < n, n + 1 < # x, x[n + 1] < x[n], x[n, x[n + 1]][2 + 2 * n0] < x[n, x[n + 1]][n + 1, x[n]][1 + 2 * n0], 2 * n0 < n, n + 1 < # x[n, x[n + 1]], ∀ n0. 2 + 2 * n0 < n → ¬ x[2 + 2 * n0] < x[1 + 2 * n0] ⊦ x[n, x[n + 1]] = [], n = 2 * n0 + 1, x = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
m < # x ⊦ m ≠ n → x[m, a][n] = x[n]
Proof step 17:
n % 2 = 1, 2 * n0 < n, x[n, x[n + 1]][2 + 2 * n0] < x[n, x[n + 1]][1 + 2 * n0], x[n + 1] < x[n], n + 1 < # x, 2 * n0 + 1 < n, n + 1 < # x[n, x[n + 1]], ∀ n0. 2 + 2 * n0 < n → ¬ x[2 + 2 * n0] < x[1 + 2 * n0] ⊦ n + 1 = 1 + 2 * n0, x[n, x[n + 1]] = [], n = 2 * n0 + 1, x = []

Rule argument

spec: list-del name: put-at seq: m < # x ⊦ m ≠ n → x[m, a][n] = x[n] subst: [n, a, m, x] → [1 + 2 * n0, x[n], n + 1, x[n, x[n + 1]]] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%n ≠ 0 ⊦ (m * n) % n = 0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: 1 + 2 * n0, x[n], n + 1, x[n, x[n + 1]]