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Proof step 3 from proof sorted-next-0-swap (previous step)

n % 2 = 0, x[n, x[n + 1]][n + 1, x[n]][1 + 2 * n0] < x[n, x[n + 1]][n + 1, x[n]][2 * n0], x[n + 1] < x[n], n + 1 < # x, 2 * n0 ≤ n, ∀ n0. 1 + 2 * n0 < n → ¬ x[1 + 2 * n0] < x[2 * n0] ⊦ x = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
m < # x ⊦ m ≠ n → x[m, a][n] = x[n]
Proof step 5:
n % 2 = 0, n + 1 < # x, x[n + 1] < x[n], x[n, x[n + 1]][n + 1, x[n]][1 + 2 * n0] < x[n, x[n + 1]][n + 1, x[n]][2 * n0], 2 * n0 ≤ n, ∀ n0. 1 + 2 * n0 < n → ¬ x[1 + 2 * n0] < x[2 * n0] ⊦ x = [], n + 1 < # x[n, x[n + 1]] ∧ n + 1 ≠ 1 + 2 * n0
Proof step 11:
n % 2 = 0, n + 1 < # x, x[n + 1] < x[n], x[n, x[n + 1]][1 + 2 * n0] < x[n, x[n + 1]][n + 1, x[n]][2 * n0], n + 1 < # x[n, x[n + 1]], 2 * n0 ≤ n, ∀ n0. 1 + 2 * n0 < n → ¬ x[1 + 2 * n0] < x[2 * n0] ⊦ n + 1 = 1 + 2 * n0, 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 terms: 1 + 2 * n0, x[n], n + 1, x[n, x[n + 1]]