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Proof step 2 from proof sorted-next-0 (previous step)

n % 2 = 0, partial-sorted(x, 0, n) ⊦ x[n + 1] < x[n], partial-sorted(x, 0, n + 2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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]);% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%n < n0 ⊦ m + (* n) < m0 + (* n0) ↔ m < m0 + (n0 - n); (s)% ⊦ n + 0 = n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic