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Proof step 1 from proof indef (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ a ∈ s ↔ (∃ s0. (∀ a0. a0 ∈ s ↔ a0 = a ∨ a0 ∈ s0) ∧ ¬ a ∈ s0)

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;