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Proof step 2 from proof suffix-max-sum-reset (previous step)

i + ar[n] < 0, n < # ar, maximum-suffix-sum(subarray(ar, 0, n), m, i) ⊦ maximum-suffix-sum(subarray(ar, 0, n + 1), n + 1, 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
sum(ar, m, n - m) + ar[n] < sum(ar, n0, n + 1 - n0), (sum(ar, m, n - m) + ar[n]) + 1 < sum(ar, n0, n + 1 - n0), m < # ar, n < # ar, sum(ar, m, n - m) + ar[n] < 0, 0 < sum(ar, n0, n + 1 - n0), suffix-upper-bound(subarray(ar, 0, n), sum(ar, m, n - m)), n0 ≤ # ar, m ≤ n, n0 ≤ n + 1, n0 ≤ n → sum(ar, n0, n - n0) ≤ sum(ar, m, n - m) ⊦

Rule argument

Additional information

Used simplifier rules: maximum-suffix-sum: ⊦ maximum-suffix-sum(ar, n, i) ↔ n ≤ # ar ∧ i = sum(ar, n, # ar - n) ∧ suffix-upper-bound(ar, i);%sum-subarray: m0 + m1 ≤ n, n ≤ # ar ⊦ sum(subarray(ar, 0, n), m0, m1) = sum(ar, m0, m1);%sum-zero: ⊦ sum(ar, n, 0) = 0;%suffix-upper-bound-reduce: ⊦ suffix-upper-bound(ar, i) → ( suffix-upper-bound(ar0, i0) ↔ (∀ n. (n ≤ # ar → sum(ar, n, # ar - n) ≤ i) → n ≤ # ar0 → sum(ar0, n, # ar0 - n) ≤ i0));% ⊦ 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 < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ i +1 = i + 1; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%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%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m - m = 0; (s)% ⊦ m + n - n + n1 = m - n1; (s)% ⊦ m < n → m ≤ n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%n ≤ m ⊦ (m - n) + m0 ≤ n0 ↔ m + m0 ≤ n0 + n; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ # subarray(ar0, m, n) = n; (s)