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Proof step 2 from proof max-sum-next-less-zero (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 + m < # ar, ar[n] < 0, n < # ar, upper-bound(subarray(ar, 0, n), sum(ar, n0, m)), n0 + m ≤ n ⊦ ∃ n1, m0. n1 + m0 ≤ n + 1 ∧ sum(ar, n0, m) = sum(ar, n1, m0) ∧ upper-bound(subarray(ar, 0, n + 1), sum(ar, n1, m0))

Rule argument

Additional information

Used simplifier rules: maximum-sum: ⊦ maximum-sum(ar, i) ↔ (∃ n, m. n + m ≤ # ar ∧ maximum-sum(ar, n, m, i));%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);% ⊦ i < 0 ∧ 0 ≤ j → i < j; (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%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ # subarray(ar0, m, n) = n; (s)