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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
sum(ar, n0, m0) < 0, sum(ar, n0, m0) < ~1, sum(ar, m, n1) + ar[n1 + m] < 0, sum(ar, n0, m0) < sum(ar, m, n1) + ar[n1 + m], n0 + m0 < # ar, m < # ar, n1 + m < # ar, upper-bound(subarray(ar, 0, n1 + m), sum(ar, n0, m0)), n0 + m0 ≤ n1 + m, ∀ n2. n2 ≤ n1 + m → sum(ar, n2, n1 + m - n2) ≤ sum(ar, m, n1) ⊦

Rule argument

Additional information

Used simplifier rules: sum-zero: ⊦ sum(ar, n, 0) = 0;% ⊦ ¬ n < n; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)%j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0); (s)% ⊦ i +1 = i + 1; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%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 → (n ≤ m ↔ false); (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (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)