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Proof step 8 from proof indef (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ a0. a0 ∈ s ↔ a0 = a ∨ a0 ∈ s0 ⊦ a ∈ s, a ∈ s0

Rule argument

Additional information