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Proof step 11 from proof dups-01 (previous step)

⊦ dups(x), ∃ a. a ∈ x ∧ (a ∈ s ∨ a ∈ s0), (¬ a ∈ s ∨ ¬ a ∈ x) ∧ (¬ a ∈ s0 ∨ ¬ a ∈ x)

User applied rule exists right

And obtained the following new goals

Proof step 12:
⊦ dups(x), a ∈ x ∧ (a ∈ s ∨ a ∈ s0), ∃ a. a ∈ x ∧ (a ∈ s ∨ a ∈ s0), (¬ a ∈ s ∨ ¬ a ∈ x) ∧ (¬ a ∈ s0 ∨ ¬ a ∈ x)

Rule argument

pos: right 2 terms: a abort: F discard: F computed: T

Additional information

Used terms: a