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Proof step 7 from proof ocs-dups-ex (
previous step
)
∃ a. a ∈ x ∧ #oc(a, x) ≠ 1 ⊦ dups(x)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 8
:
a ∈ x ⊦ #oc(a, x) = 1, dups(x)
Rule argument
Additional information