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Proof step 6 from proof diff-dups (previous step)

a ∈ y, dups(x0 \ y), ∀ y0. ¬ dups(x0 \ y0) ⊦ a ∈ x0, dups(x0)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: y

Additional information

Used terms: y