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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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);%sum-succ: n + m < # ar ⊦ sum(ar, n, m + 1) = sum(ar, n, m) + ar[n + m];%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ n < m → (m - n) + n = m; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m ≤ n → (n - m = 0 ↔ n = m); (s)% ⊦ m < m + n ↔ n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ 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)% ⊦ n ≤ n0 → (m + (n0 - n)) + n = m + n0; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (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)