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Proof step 1 from proof range-dups (conclusion)

∀ i. i ∈ ints → 0 ≤ i ∧ i < j, 0 ≤ j, j < n→i(# ints) ⊦ dups(ints)

User applied rule induction

And obtained the following new goals

Proof step 2:
(∀ i. i ∈ ints → 0 ≤ i ∧ i < j) ∧ 0 ≤ j ∧ j < n→i(# ints), ∀ j0, ints. i→n(j0) < i→n(j) → (∀ i. i ∈ ints → 0 ≤ i ∧ i < j0) ∧ 0 ≤ j0 ∧ j0 < n→i(# ints) → dups(ints) ⊦ dups(ints)

Rule argument

indvar: i→n(j) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j)