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Proof step 2 from proof sorted-next-1-swap (previous step)

n % 2 = 1, x[n + 1] < x[n], n + 1 < # x, partial-sorted(x, 1, n) ⊦ partial-sorted(x[n, x[n + 1]][n + 1, x[n]], 1, n + 2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n % 2 = 1, 2 * n0 < n, x[n, x[n + 1]][n + 1, x[n]][2 + 2 * n0] < x[n, x[n + 1]][n + 1, x[n]][1 + 2 * n0], x[n + 1] < x[n], n + 1 < # x, ∀ n0. 2 + 2 * n0 < n → ¬ x[2 + 2 * n0] < x[1 + 2 * n0] ⊦ x = []

Rule argument

Additional information

Used simplifier rules: partial-sorted: ⊦ partial-sorted(x, n, m) ↔ (∀ n0. n + 2 * n0 + 1 < m → ¬ x[n + 2 * n0 + 1] < x[n + 2 * n0]);% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic