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Proof step 6 from proof suffix-max-sum-reset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(n2 + m) + 1 = n1 + n0, sum(ar, m, n2) + ar[n2 + m] < sum(ar, n0, n1), (sum(ar, m, n2) + ar[n2 + m]) + 1 < sum(ar, n0, n1), m < # ar, n2 + m < # ar, sum(ar, m, n2) + ar[n2 + m] < 0, 0 < sum(ar, n0, n1), suffix-upper-bound(subarray(ar, 0, n2 + m), sum(ar, m, n2)), n0 ≤ # ar, n1 + n0 ≤ # ar, n0 ≤ n2 + m → sum(ar, n0, n2 + m - n0) ≤ sum(ar, m, n2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (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)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%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%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)