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Proof step 2 from proof sorted-shift-one (previous step)

m + 1 < # A, m + 1 < n0, sorted(A, n, n0), n ≤ m ⊦ sorted(A[m + 1, A[m]], n, n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n1 + 1 < # A, n1 + 1 < n0, A[m + 1, A[m]][n1 + 1] < A[m + 1, A[m]][n1], m + 1 < # A, m + 1 < n0, n ≤ n1, n0 ≤ # A, n ≤ m, ∀ n1. n ≤ n1 ∧ n1 + 1 < n0 → ¬ 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% ⊦ # ar[n, a] = # ar; (s)