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Proof step 15 from proof suffix-upper-bound-reduce (previous step)

# ar0 = n0 + n, i0 < sum(ar0, n, n0), ∀ n. n ≤ # ar → sum(ar, n, # ar - n) ≤ i, ∀ n. (n ≤ # ar → sum(ar, n, # ar - n) ≤ i) → n ≤ # ar0 → sum(ar0, n, # ar0 - n) ≤ i0 ⊦

User applied rule all left

And obtained the following new goals

Proof step 16:
# ar0 = n0 + n, i0 < sum(ar0, n, n0), (n ≤ # ar → sum(ar, n, # ar - n) ≤ i) → n ≤ # ar0 → sum(ar0, n, # ar0 - n) ≤ i0, ∀ n. (n ≤ # ar → sum(ar, n, # ar - n) ≤ i) → n ≤ # ar0 → sum(ar0, n, # ar0 - n) ≤ i0, ∀ n. n ≤ # ar → sum(ar, n, # ar - n) ≤ i ⊦

Rule argument

pos: left 4 terms: n abort: F discard: F computed: F

Additional information

Used terms: n