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Proof step 17 from proof sorted-next-0-swap (previous step)

n % 2 = 0, x[1 + 2 * n0] < x[2 * n0], x[n + 1] < x[n], n + 1 < # x, 2 * n0 < n, n + 1 < # x[n, x[n + 1]], n < # x, ∀ n0. 1 + 2 * n0 < n → ¬ x[1 + 2 * n0] < x[2 * n0] ⊦ n + 1 = 2 * n0, x[n, x[n + 1]] = [], n = 2 * n0, n = 1 + 2 * n0, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
n % 2 = 0, 2 * n0 + 1 < # x, 2 * n0 < n, x[n + 1] < x[n], x[1 + 2 * n0] < x[2 * n0], n < # x, n + 1 < # x, 2 * n0 < # x, 2 * n0 + 1 < n, 2 + 2 * n0 < # x, ∀ n0. 1 + 2 * n0 < n → ¬ x[1 + 2 * n0] < x[2 * n0] ⊦ n + 1 = 2 * n0, x[n, x[n + 1]] = [], n = 1 + 2 * n0, n = 2 * n0, x = []

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic% ⊦ n < # x → # x[n, a] = # x; (s)