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

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

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] → [n1, n2 + m, n3]

Additional information

Used simplifier rules: lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%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% ⊦ i < j → i1 + i < i1 + j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ m ≤ m + n; (s)% ⊦ i < j → (j ≤ i ↔ false); (as)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n1, n2 + m, n3