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Proof step 14 from proof sorted-next-1-swap (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
n % 2 = 1, 2 * n0 + 1 < n, n + 1 < # x, x[n + 1] < x[n], x[n, x[n + 1]][2 + 2 * n0] < x[n, x[n + 1]][n + 1, x[n]][1 + 2 * n0], 2 * n0 < n, n + 1 < # x[n, x[n + 1]], ∀ n0. 2 + 2 * n0 < n → ¬ x[2 + 2 * n0] < x[1 + 2 * n0] ⊦ x[n, x[n + 1]] = [], n = 2 * n0 + 1, x = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (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%n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s)