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Proof step 1 from proof oc-diff (conclusion)

⊦ a ∈ y → #oc(a, x \ y) = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%notin-01: ⊦ a ∈ y → ¬ a ∈ x \ y; from specification list-set