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Proof step 6 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 = [], n + 1 < # x[n, x[n + 1]] ∧ n ≠ 2 * n0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ n < # x → # x[n, a] = # x
Proof step 8:
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], n < # x, 2 * n0 ≤ n, ∀ n0. 1 + 2 * n0 < n → ¬ x[1 + 2 * n0] < x[2 * n0] ⊦ x = [], n + 1 < # x ∧ n ≠ 2 * n0

Rule argument

spec: list-del name: put-length seq: ⊦ n < # x → # x[n, a] = # x subst: [a, n, x] → [x[n + 1], n, x] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ m < n → ¬ m + m0 < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: x[n + 1], n, x