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Proof step 10 from proof max-sum-keep-current-max (previous step)

n1 + m1 = n + 1, n = n2 + m, n < # ar, sum(ar, m, n2) + ar[n] < sum(ar, n1, m1), n0 + m0 < # ar, m < # ar, sum(ar, n0, m0) < sum(ar, n1, m1), suffix-upper-bound(subarray(ar, 0, n), sum(ar, m, n2)), upper-bound(subarray(ar, 0, n), sum(ar, n0, m0)), n0 + m0 ≤ n + 1, sum(ar, m, n2) + ar[n] ≤ sum(ar, n0, m0), n0 + m0 ≤ n, m ≤ n ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n1 + m1 = (n2 + m) + 1, sum(ar, n0, m0) < sum(ar, n1, m1), m < # ar, n0 + m0 < # ar, n2 + m < # ar, sum(ar, m, n2) + ar[n2 + m] < sum(ar, n1, m1), suffix-upper-bound(subarray(ar, 0, n2 + m), sum(ar, m, n2)), upper-bound(subarray(ar, 0, n2 + m), sum(ar, n0, m0)), n0 + m0 ≤ n2 + m, sum(ar, m, n2) + ar[n2 + m] ≤ sum(ar, n0, m0), n0 + m0 ≤ n1 + m1 ⊦

Rule argument

Additional information

Used simplifier rules: forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)