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Proof step 5 from proof suffix-max-sum-reset (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
n + 1 = n1 + n0, n = n2 + m, (sum(ar, m, n2) + ar[n]) + 1 < sum(ar, n0, n1), m < # ar, n < # ar, sum(ar, m, n2) + ar[n] < 0, 0 < sum(ar, n0, n1), sum(ar, m, n2) + ar[n] < sum(ar, n0, n1), suffix-upper-bound(subarray(ar, 0, n), sum(ar, m, n2)), n0 ≤ # ar, m ≤ n, n1 + n0 ≤ # ar, n0 ≤ n → sum(ar, n0, n - n0) ≤ sum(ar, m, n2) ⊦

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