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Proof step 2 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), i), j + ar[n] ≤ i ⊦ maximum-sum(subarray(ar, 0, n + 1), i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ maximum-sum(ar, i) ↔ (∃ n, m. n + m ≤ # ar ∧ maximum-sum(ar, n, m, i))
Proof step 4:
n < # ar, maximum-suffix-sum(subarray(ar, 0, n), m, j), maximum-sum(subarray(ar, 0, n), i), j + ar[n] ≤ i ⊦ ∃ n0, m. n0 + m ≤ # subarray(ar, 0, n + 1) ∧ maximum-sum(subarray(ar, 0, n + 1), n0, m, i)

Rule argument

spec: <toplevel> name: maximum-sum seq: ⊦ maximum-sum(ar, i) ↔ (∃ n, m. n + m ≤ # ar ∧ maximum-sum(ar, n, m, i)) subst: [i, ar] → [i, subarray(ar, 0, n + 1)] flags: <all paths>

Additional information

Used terms: i, subarray(ar, 0, n + 1)