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Proof step 2 from proof sorted-merge (previous step)

m > n0, sorted(A, n, m), sorted(A, n0, m0) ⊦ sorted(A, n, m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 < m, A[n1 + 1] < A[n1], n1 + 1 < m0, n0 < # A, n1 + 1 < # A, m0 ≤ # A, m ≤ # A, n ≤ n1, ∀ n1. n0 ≤ n1 ∧ n1 + 1 < m0 → ¬ A[n1 + 1] < A[n1], ∀ n1. n ≤ n1 ∧ n1 + 1 < m → ¬ A[n1 + 1] < A[n1] ⊦

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% ⊦ m > n ↔ n < m; (s)