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Proof step 6 from proof indef (previous step)

a ∈ s, ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 ⊦ ∃ s0. (∀ a0. a0 ∈ s ↔ a0 = a ∨ a0 ∈ s0) ∧ ¬ a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ a ∈ s0, ∃ s. (∀ a0. a0 ≠ a → (a0 ∈ s0 ↔ a0 ∈ s)) ∧ ¬ a ∈ s

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;