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Proof step 7 from proof max-sum-keep-current-max (previous step)

n < # ar, maximum-suffix-sum(subarray(ar, 0, n), m, j), maximum-sum(subarray(ar, 0, n), n0, m0, i), j + ar[n] ≤ i, n0 + m0 ≤ # subarray(ar, 0, n) ⊦ ∃ n0, m. n0 + m ≤ # subarray(ar, 0, n + 1) ∧ maximum-sum(subarray(ar, 0, n + 1), n0, m, i)

User applied rule exists right

And obtained the following new goals

Proof step 8:
n < # ar, maximum-suffix-sum(subarray(ar, 0, n), m, j), maximum-sum(subarray(ar, 0, n), n0, m0, i), n0 + m0 ≤ # subarray(ar, 0, n), j + ar[n] ≤ i ⊦ n0 + m0 ≤ # subarray(ar, 0, n + 1) ∧ maximum-sum(subarray(ar, 0, n + 1), n0, m0, i)

Rule argument

pos: right 1 terms: n0, m0 abort: F discard: T computed: F

Additional information

Used terms: n0, m0