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Proof step 3 from proof max-sum-next-less-zero (previous step)

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))

User applied rule exists right

And obtained the following new goals

Proof step 4:
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))

Rule argument

pos: right 1 terms: n0, m abort: F discard: T computed: F

Additional information

Used terms: n0, m