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Proof step 2 from proof sorted-lower-higher-bound (previous step)

sorted(A, n, m0), m ≤ m0 ⊦ sorted(A, n, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
A[n0 + 1] < A[n0], n0 + 1 < m, n0 + 1 < # A, n0 + 1 < m0, m ≤ # A, m0 ≤ # A, m ≤ m0, n ≤ n0, ∀ n0. n ≤ n0 ∧ n0 + 1 < m0 → ¬ A[n0 + 1] < A[n0] ⊦

Rule argument

Additional information

Used simplifier rules: sorted: ⊦ sorted(ar, n, m) ↔ m ≤ # ar ∧ (∀ n0. n ≤ n0 ∧ n0 + 1 < m → ar[n0] ≤ ar[n0 + 1]);%less-eq-elem: ⊦ x ≤ y ↔ ¬ y < x;% ⊦ 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%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat