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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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