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Proof step 10 from proof upper-bound-reduce (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ upper-bound(ar, i) ↔ (∀ n, m. n + m ≤ # ar → sum(ar, n, m) ≤ i)
Proof step 12:
i0 < sum(ar0, n, m), n + m ≤ # ar0, ∀ n, m. (n + m ≤ # ar → sum(ar, n, m) ≤ i) → n + m ≤ # ar0 → sum(ar0, n, m) ≤ i0, ∀ n, m. n + m ≤ # ar → sum(ar, n, m) ≤ i ⊦

Rule argument

spec: <toplevel> name: upper-bound seq: ⊦ upper-bound(ar, i) ↔ (∀ n, m. n + m ≤ # ar → sum(ar, n, m) ≤ i) subst: [i, ar] → [i, ar] flags: <all paths>

Additional information

Used terms: i, ar