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Proof step 17 from proof suffix-upper-bound-reduce (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
# ar0 = n0 + n, # ar = n1 + n, i < sum(ar, n, n1), i0 < sum(ar0, n, n0), n ≤ # ar, ∀ 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 ⊦

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [n, # ar, n1]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # ar, n1