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Proof step 5 from proof sorted-next-0-swap (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)