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Proof step 13 from proof max-sum-new-current-max (previous step)

n + m1 = (n1 + m) + 1, n1 + m < # ar, n0 + m0 < n + m1, sum(ar, m, n1) + ar[n1 + m] < sum(ar, n, m1), sum(ar, n0, m0) < sum(ar, n, m1), n1 + m < n + m1, sum(ar, n0, m0) + 1 < sum(ar, n, m1), m < # ar, n0 + m0 < # ar, sum(ar, n0, m0) < sum(ar, m, n1) + ar[n1 + m], upper-bound(subarray(ar, 0, n1 + m), sum(ar, n0, m0)), n + m1 ≤ # ar, n0 + m0 ≤ n1 + m, ∀ n. n ≤ # subarray(ar, 0, n1 + m) → sum(subarray(ar, 0, n1 + m), n, # subarray(ar, 0, n1 + m) - n) ≤ sum(ar, m, n1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
n + m1 = (n1 + m) + 1, sum(ar, n0, m0) < sum(ar, m, n1) + ar[n1 + m], n0 + m0 < # ar, m < # ar, sum(ar, n0, m0) + 1 < sum(ar, n, m1), n1 + m < n + m1, sum(ar, n0, m0) < sum(ar, n, m1), sum(ar, m, n1) + ar[n1 + m] < sum(ar, n, m1), n0 + m0 < n + m1, n1 + m < # ar, upper-bound(subarray(ar, 0, n1 + m), sum(ar, n0, m0)), n0 + m0 ≤ n1 + m, n + m1 ≤ # ar, ∀ n2. n2 ≤ n1 + m → sum(ar, n2, n1 + m - n2) ≤ sum(ar, m, n1) ⊦

Rule argument

Additional information

Used simplifier rules: sum-subarray: m0 + m1 ≤ n, n ≤ # ar ⊦ sum(subarray(ar, 0, n), m0, m1) = sum(ar, m0, m1);% ⊦ m < n → m ≤ n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%n ≤ m ⊦ (m - n) + m0 ≤ n0 ↔ m + m0 ≤ n0 + n; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ # subarray(ar0, m, n) = n; (s)