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Proof step 19 from proof max-sum-new-current-max (previous step)

n1 + m < # ar, m < # ar, n0 + m0 < # ar, sum(ar, n0, m0) < sum(ar, m, n1) + ar[n1 + m], sum(ar, m, n1) + ar[n1 + m] < 0, sum(ar, n0, m0) < ~1, sum(ar, n0, m0) < 0, n0 + m0 ≤ n1 + m, ∀ n2. n2 ≤ n1 + m → sum(ar, n2, n1 + m - n2) ≤ sum(ar, m, n1), ∀ n, m1. n + m1 ≤ # subarray(ar, 0, n1 + m) → sum(subarray(ar, 0, n1 + m), n, m1) ≤ sum(ar, n0, m0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
sum(ar, n0, m0) < 0, sum(ar, n0, m0) < ~1, sum(ar, m, n1) + ar[n1 + m] < 0, sum(ar, n0, m0) < sum(ar, m, n1) + ar[n1 + m], n0 + m0 < # ar, m < # ar, n1 + m < # ar, n0 + m0 ≤ n1 + m, ∀ n2. n2 ≤ n1 + m → sum(ar, n2, n1 + m - n2) ≤ sum(ar, m, n1), ∀ n, m1. n + m1 ≤ n1 + m → sum(ar, n, m1) ≤ sum(ar, n0, m0) ⊦

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)% ⊦ # subarray(ar0, m, n) = n; (s)