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Proof step 2 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)

User applied rule case distinction

And obtained the following new goals

Proof step 3:
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
Proof step 7:
upper-bound(ar, i), ∀ n, m. (n + m ≤ # ar → sum(ar, n, m) ≤ i) → n + m ≤ # ar0 → sum(ar0, n, m) ≤ i0 ⊦ upper-bound(ar0, i0)

Rule argument

pos: right 1

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