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Proof step 1 from proof oc-zero (conclusion)

⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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