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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # ar, m < # ar, suffix-upper-bound(subarray(ar, 0, n), sum(ar, m, n - m)), m ≤ n + 1, 0 ≤ sum(ar, m, n - m) + ar[n], 0 ≤ sum(ar, m, n - m), m ≤ n ⊦ suffix-upper-bound(subarray(ar, 0, n + 1), sum(ar, m, n - m) + ar[n]) ∧ ar[n] = 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-succ: n + m < # ar ⊦ sum(ar, n, m + 1) = sum(ar, n, m) + ar[n + m];% ⊦ ¬ m + n < m; (s)% ⊦ m +1 = m + 1; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%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)%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%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)% ⊦ 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)