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Proof step 4 from proof suffix-max-sum-reset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n + 1 = n1 + n0, sum(ar, m, n - m) + ar[n] < sum(ar, n0, n1), 0 < sum(ar, n0, n1), sum(ar, m, n - m) + ar[n] < 0, n < # ar, m < # ar, (sum(ar, m, n - m) + ar[n]) + 1 < sum(ar, n0, n1), suffix-upper-bound(subarray(ar, 0, n), sum(ar, m, n - m)), n1 + n0 ≤ # ar, m ≤ n, n0 ≤ # ar, n0 ≤ n → sum(ar, n0, n - n0) ≤ sum(ar, m, n - m) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (s)% ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i +1 = i + 1; (s)%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (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%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat