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Proof step 17 from proof sorted-next-1-swap (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
n % 2 = 1, 2 * n0 < n, n + 1 < # x[n, x[n + 1]], 2 * n0 + 1 < n, n + 1 < # x, x[n + 1] < x[n], x[n, x[n + 1]][2 + 2 * n0] < x[1 + 2 * n0], ∀ n0. 2 + 2 * n0 < n → ¬ x[2 + 2 * n0] < x[1 + 2 * n0] ⊦ x[n, x[n + 1]] = [], n = 2 * n0 + 1, n = 2 * n0, x = []

Rule argument

Additional information

Used simplifier rules: m < # x ⊦ m ≠ n → x[m, a][n] = x[n]; (s)% ⊦ ¬ m + n < m; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)%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)