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Proof step 7 from proof nodups (previous step)

#oc(a, x) = 1, #oc(a, y) = 0, ∀ a. a ∈ x ↔ a ∈ y ⊦ dups(y), dups(x)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a

Additional information

Used simplifier rules: notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0; Used terms: a