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Proof step 2 from proof ocnodups-01 (previous step)

⊦ 1 ≤ #oc(a, x) ↔ a ∈ x, dups(x)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)