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Proof step 11 from proof ocs-dups-ex (previous step)

a = a0 ∨ a ∈ x0, ∀ a1. a1 ∈ x0 → #oc(a1, x0) = 1 ⊦ a = a0, a0 ∈ x0, dups(x0), a ≠ a0 ∧ #oc(a, x0) = 1, ¬ a ∈ x0 ∧ #oc(a, a0 ') = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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