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Proof step 15 from proof max-sum-next-less-zero (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: n1, m1

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m ≤ m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < i + j ↔ 0 < j; (s)% ⊦ i < j → ¬ j < i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: n1, m1