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

sorted(ar, n + 1, m), a ≤ ar[n + 1] ⊦ sorted(ar[n, a], n, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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 < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%m ≠ n ⊦ ar[m, a][n] = ar[n]; (s)% ⊦ ¬ n < n; (s)% ⊦ # ar[n, a] = # ar; (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)