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Proof step 8 from proof sorted-next-1-swap (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem% ⊦ n0 < m0 → (m0 * m + n0 < m0 * n ↔ m < n); (s)%n0 < n ⊦ m + (* n) < m0 + (* n0) ↔ m + (n - n0) < m0; (s)%m < # x ⊦ x[m, a][m] = a; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ n < # x → # x[n, a] = # x; (s)% ⊦ ¬ n < n; (s)% ⊦ 2 * m < 2 * n ↔ m < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ (2 * m + 1) % 2 = 1; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (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%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat