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Proof step 4 from proof max-sum-next-less-zero (previous step)

n < # ar, ar[n] < 0, n0 + m < # ar, upper-bound(subarray(ar, 0, n), sum(ar, n0, m)), n0 + m ≤ n ⊦ n0 + m ≤ n + 1 ∧ sum(ar, n0, m) = sum(ar, n0, m) ∧ upper-bound(subarray(ar, 0, n + 1), sum(ar, n0, m))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: 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));%sum-subarray: m0 + m1 ≤ n, n ≤ # ar ⊦ sum(subarray(ar, 0, n), m0, m1) = sum(ar, m0, m1);% ⊦ ¬ n < n; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%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%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 + 1 ≤ n ↔ m < n; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ # subarray(ar0, m, n) = n; (s)% ⊦ ¬ m + n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (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%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)