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Proof step 11 from proof max-sum-next-less-zero (previous step)

n1 + m0 = n + 1, m0 = 0, n < # ar, n0 + m < # ar, ar[n] < 0, sum(ar, n0, m) < sum(ar, n1, m0), n0 + m ≤ n + 1, n0 + m ≤ n, ∀ n2, m1. n2 + m1 ≤ n → sum(ar, n2, m1) ≤ sum(ar, n0, m) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
sum(ar, n0, m) < 0, n0 + m < # ar, ar[n] < 0, n < # ar, n0 + m ≤ n + 1, n0 + m ≤ n, ∀ n2, m1. n2 + m1 ≤ n → sum(ar, n2, m1) ≤ sum(ar, n0, m) ⊦

Rule argument

Additional information

Used simplifier rules: sum-zero: ⊦ sum(ar, n, 0) = 0;% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%eqle-one: ⊦ m = n + 1 ∧ 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% ⊦ n + 0 = n; (s)