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Proof step 10 from proof nodups-butlastn (previous step)

dups(butlastn(# x0, a ' + x0.butlast)), ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ x0 = [], dups(x0), a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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