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Proof step 12 from proof ocs-dups-ex (previous step)

a ∈ x0, ∀ a1. a1 ∈ x0 → #oc(a1, x0) = 1 ⊦ #oc(a, x0) = 1, a0 ∈ x0, dups(x0)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a

Additional information

Used terms: a