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Proof step 4 from proof notin (previous step)

∀ a. #oc(a, x0) = 0 ∨ a ∈ x0 ⊦ a = a0, #oc(a, x0) = 0, a ∈ x0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: a

Additional information

Used terms: a