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Proof step 1 from proof indef (conclusion)

⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)

User applied rule case distinction

And obtained the following new goals

Proof step 2:
a ∈ s ⊦ ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0

Rule argument

pos: right 1

Additional information

Used simplifier rules: in-03: ⊦ ¬ a ∈ s → (a ∈ s ++ b ↔ a = b); from specification infset-basic