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Proof step 2 from proof sorted-inc-right (previous step)

m0 = m + 1, m < # ar, sorted(ar, n, m), ar[m - 1] ≤ a ⊦ sorted(ar[m, a], n, m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 + 1 < # ar, m < # ar, ar[m, a][n0 + 1] < ar[n0], n0 < m, n < m, n0 < # ar, n < # ar, n + 1 < # ar, n ≤ n0, ∀ n0. n ≤ n0 ∧ n0 + 1 < m → ¬ ar[n0 + 1] < ar[n0] ⊦ a < ar[m - 1]

Rule argument

Additional information

Used simplifier rules: less-eq-elem: ⊦ x ≤ y ↔ ¬ y < x;%sorted: ⊦ sorted(ar, n, m) ↔ m ≤ # ar ∧ (∀ n0. n ≤ n0 ∧ n0 + 1 < m → ar[n0] ≤ ar[n0 + 1]);% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ n < m → ar[m, a][n] = ar[n]; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ # ar[n, a] = # ar; (s)% ⊦ m < n → m ≤ n; (s)