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Proof step 7 from proof indef (previous step)

⊦ a ∈ s0, ∃ s. (∀ a0. a0 ≠ a → (a0 ∈ s0 ↔ a0 ∈ s)) ∧ ¬ a ∈ s

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: s0 abort: F discard: T computed: T

Additional information

Used terms: s0