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Proof step 1 from proof ocs-dups-ex (conclusion)

⊦ dups(x) ↔ (∃ a. a ∈ x ∧ #oc(a, x) > 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ dups(x) ↔ (∃ a. a ∈ x ∧ #oc(a, x) ≠ 1)

Rule argument

Additional information

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