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Proof step 22 from proof max-sum-new-current-max (previous step)

n + m2 = n1 + m, sum(ar, n0, m0) < sum(ar, m, n1) + ar[n1 + m], n0 + m0 < # ar, m < # ar, n1 + m < # ar, sum(ar, n0, m0) + 1 < sum(ar, n, m2) + ar[n1 + m], sum(ar, m, n1) < sum(ar, n, m2), n < # ar, sum(ar, n, n1 + m - n) < sum(ar, n, m2), upper-bound(subarray(ar, 0, n1 + m), sum(ar, n0, m0)), n0 + m0 ≤ n1 + m, n ≤ n1 + m, sum(ar, n, n1 + m - n) ≤ sum(ar, m, n1), ∀ n2. n2 ≤ n1 + m → sum(ar, n2, n1 + m - n2) ≤ sum(ar, m, n1) ⊦

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [n, n1 + m, n3]

Additional information

Used simplifier rules: forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ m + n; (s)% ⊦ i < j → (j ≤ i ↔ false); (as)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, n1 + m, n3