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

0 < m, sorted(A, 0, m), sorted(A, m + 1, n), A[m - 1] ≤ a, a ≤ A[m + 1] ⊦ sorted(A[m, a], 0, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
A[m, a][n0 + 1] < A[m, a][n0], n0 + 1 < n, n0 + 1 < # A, n ≤ # A, m ≤ # A, ∀ n0. m < n0 ∧ n0 + 1 < n → ¬ A[n0 + 1] < A[n0], ∀ n0. n0 + 1 < m → ¬ A[n0 + 1] < A[n0] ⊦ m = 0, A[m + 1] < a, a < A[m - 1], A[m + 1] < A[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 < n → n ≠ 0; (s)%nnn: ⊦ ¬ c < b ∧ ¬ a < c → ¬ a < b; (forward) from specification oelem%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)% ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ 0 ≤ m; (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)