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Proof step 21 from proof max-sum-new-current-max (previous step)

n + m1 = (n1 + m) + 1, m1 = m2 +1, 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 22:
n + m2 = n1 + m, sum(ar, n0, m0) < sum(ar, m, n1) + ar[n1 + m], n0 + m0 < # ar, m < # ar, n1 + m < # ar, sum(ar, n0, m0) + 1 < sum(ar, n, m2) + ar[n1 + m], sum(ar, m, n1) < sum(ar, n, m2), n < # ar, sum(ar, n, n1 + m - n) < sum(ar, n, m2), upper-bound(subarray(ar, 0, n1 + m), sum(ar, n0, m0)), n0 + m0 ≤ n1 + m, 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) ⊦

Rule argument

Additional information

Used simplifier rules: sum-succ: n + m < # ar ⊦ sum(ar, n, m + 1) = sum(ar, n, m) + ar[n + m];%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ ¬ m + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ i < j + i0 ∧ j < j0 → i < j0 + i0; (s)% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ m ≤ m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (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%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)