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Proof step 2 from proof ocs-dups-ex (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 3:
dups(x) ⊦ ∃ a. a ∈ x ∧ #oc(a, x) ≠ 1
Proof step 7:
∃ a. a ∈ x ∧ #oc(a, x) ≠ 1 ⊦ dups(x)

Rule argument

pos: right 1

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