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Proof step 11 from proof sorted-next-0-swap (previous step)

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 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n % 2 = 0, 2 * n0 < n, 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]], ∀ n0. 1 + 2 * n0 < n → ¬ x[1 + 2 * n0] < x[2 * n0] ⊦ x[n, x[n + 1]] = [], n = 2 * n0, x = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)%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)