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Proof step 30 from proof nodups-butlastn (previous step)

dups(butlastn(n, x0)), n ≤ # x0, n ≤ 1 + # x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ dups(x0), a ∈ butlastn(n, x0), a ∈ x0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: n

Additional information

Used terms: n