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Proof step 19 from proof suffix-upper-bound-reduce (previous step)

# ar0 = n0 + n, # ar = n1 + n, i < sum(ar, n, n1), 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 closed the goal!

Rule argument

pos: left 5 terms: n abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n - m = n; (s)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: n