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Proof step 16 from proof max-sum-keep-current-max (previous step)

n1 + m1 = (n2 + m) + 1, m1 = 0, 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), 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, n1 ≤ n2 + m → sum(ar, n1, n2 + m - n1) ≤ sum(ar, m, n2), ∀ n. n ≤ n2 + m → sum(ar, n, n2 + m - n) ≤ sum(ar, m, n2) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
sum(ar, m, n2) + ar[n2 + m] < 0, sum(ar, n0, m0) < 0, m < # ar, n0 + m0 < # ar, n2 + m < # ar, 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 ≤ (n2 + m) + 1, ∀ n. n ≤ n2 + m → sum(ar, n, n2 + m - n) ≤ sum(ar, m, n2) ⊦

Rule argument

Additional information

Used simplifier rules: sum-zero: ⊦ sum(ar, n, 0) = 0;% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ m = n + 1 → ¬ m ≤ n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)