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Proof step 29 from proof nodups-butlastn (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: n

Additional information

Used terms: n