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Proof step 12 from proof dups-01 (
previous step
)
⊦ dups(x), a ∈ x ∧ (a ∈ s ∨ a ∈ s0), ∃ a. a ∈ x ∧ (a ∈ s ∨ a ∈ s0), (¬ a ∈ s ∨ ¬ a ∈ x) ∧ (¬ a ∈ s0 ∨ ¬ a ∈ x)
User applied rule case distinction
And closed the goal!
Rule argument
pos: right 4
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