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Proof step 8 from proof sum-all-elems-le (previous step)

Σ ints0 ≤ i0 * n→i(# ints0), i0 * (1 + n→i(# ints0)) < i + Σ ints0, ∀ i1. i1 = i ∨ i1 ∈ ints0 → i1 ≤ i0 ⊦

User applied rule all left

And obtained the following new goals

Proof step 9:
Σ ints0 ≤ i0 * n→i(# ints0), i0 * (1 + n→i(# ints0)) < i + Σ ints0, i = i ∨ i ∈ ints0 → i ≤ i0, ∀ i1. i1 ∈ ints0 → i1 ≤ i0 ⊦

Rule argument

pos: left 3 terms: i abort: F discard: F computed: T

Additional information

Used terms: i