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Proof step 9 from proof max-sum-keep-current-max (previous step)

n1 + m1 = n + 1, sum(ar, n0, m0) < sum(ar, n1, m1), m < # ar, n0 + m0 < # ar, sum(ar, m, n - m) + ar[n] < sum(ar, n1, m1), n < # ar, suffix-upper-bound(subarray(ar, 0, n), sum(ar, m, n - m)), upper-bound(subarray(ar, 0, n), sum(ar, n0, m0)), m ≤ n, n0 + m0 ≤ n, sum(ar, m, n - m) + ar[n] ≤ sum(ar, n0, m0), n0 + m0 ≤ n + 1 ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
n1 + m1 = n + 1, n = n2 + m, n < # ar, sum(ar, m, n2) + ar[n] < sum(ar, n1, m1), n0 + m0 < # ar, m < # ar, sum(ar, n0, m0) < sum(ar, n1, m1), suffix-upper-bound(subarray(ar, 0, n), sum(ar, m, n2)), upper-bound(subarray(ar, 0, n), sum(ar, n0, m0)), n0 + m0 ≤ n + 1, sum(ar, m, n2) + ar[n] ≤ sum(ar, n0, m0), n0 + m0 ≤ n, m ≤ n ⊦

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [m, n, n2]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m, n, n2