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Proof step 8 from proof max-sum-keep-current-max (previous step)

n < # ar, maximum-suffix-sum(subarray(ar, 0, n), m, j), maximum-sum(subarray(ar, 0, n), n0, m0, i), n0 + m0 ≤ # subarray(ar, 0, n), j + ar[n] ≤ i ⊦ n0 + m0 ≤ # subarray(ar, 0, n + 1) ∧ maximum-sum(subarray(ar, 0, n + 1), n0, m0, i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n1 + m1 = n + 1, sum(ar, n0, m0) < sum(ar, n1, m1), m < # ar, n0 + m0 < # ar, sum(ar, m, n - m) + ar[n] < sum(ar, n1, m1), n < # ar, suffix-upper-bound(subarray(ar, 0, n), sum(ar, m, n - m)), upper-bound(subarray(ar, 0, n), sum(ar, n0, m0)), m ≤ n, n0 + m0 ≤ n, sum(ar, m, n - m) + ar[n] ≤ sum(ar, n0, m0), n0 + m0 ≤ n + 1 ⊦

Rule argument

Additional information

Used simplifier rules: maximum-sum-bounds: ⊦ maximum-sum(ar, n, m, i) ↔ n + m ≤ # ar ∧ i = sum(ar, n, m) ∧ upper-bound(ar, i);%sum-subarray: m0 + m1 ≤ n, n ≤ # ar ⊦ sum(subarray(ar, 0, n), m0, m1) = sum(ar, m0, m1);%maximum-suffix-sum: ⊦ maximum-suffix-sum(ar, n, i) ↔ n ≤ # ar ∧ i = sum(ar, n, # ar - n) ∧ suffix-upper-bound(ar, i);%upper-bound-reduce: ⊦ upper-bound(ar, i) → ( upper-bound(ar0, i0) ↔ (∀ n, m. (n + m ≤ # ar → sum(ar, n, m) ≤ i) → n + m ≤ # ar0 → sum(ar0, n, m) ≤ i0));% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ ¬ m + n < m; (s)% ⊦ m +1 = m + 1; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m + 1 ≤ 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)% ⊦ m < n → m ≤ n; (s)% ⊦ # subarray(ar0, m, n) = n; (s)