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Proof step 3 from proof upper-bound-reduce (previous step)

upper-bound(ar, i), upper-bound(ar0, i0) ⊦ ∀ n, m. (n + m ≤ # ar → sum(ar, n, m) ≤ i) → n + m ≤ # ar0 → sum(ar0, n, m) ≤ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i0 < sum(ar0, n, m), upper-bound(ar0, i0), upper-bound(ar, i), n + m ≤ # ar0, n + m ≤ # ar → sum(ar, n, m) ≤ i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss)