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

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [m, n, n1]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m, n, n1