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Proof step 11 from proof range-dups (previous step)

i ∈ ints, j < n→i(# ints), j ≤ n→i(# ints), 0 < j, ∀ i. i ∈ ints → 0 ≤ i ∧ i < j, ∀ j0, ints. i→n(j0) < i→n(j) → (∀ i. i ∈ ints → 0 ≤ i ∧ i < j0) ∧ 0 ≤ j0 ∧ j0 < n→i(# ints) → dups(ints) ⊦ ints = [], j = 0, # ints = 1, dups(ints), 0 ≤ i ∧ i + 1 < j

User applied rule all left

And obtained the following new goals

Proof step 12:
i ∈ ints, j < n→i(# ints), j ≤ n→i(# ints), 0 < j, i ∈ ints → 0 ≤ i ∧ i < j, ∀ i. i ∈ ints → 0 ≤ i ∧ i < j, ∀ j0, ints. i→n(j0) < i→n(j) → (∀ i. i ∈ ints → 0 ≤ i ∧ i < j0) ∧ 0 ≤ j0 ∧ j0 < n→i(# ints) → dups(ints) ⊦ ints = [], j = 0, # ints = 1, dups(ints), 0 ≤ i ∧ i + 1 < j

Rule argument

pos: left 5 terms: i abort: F discard: F computed: T

Additional information

Used terms: i