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Proof step 1 from proof ext-rec (conclusion)

⊦ s1 = s0 ++ a ↔ (∀ b. b ∈ s1 ↔ b = a ∨ b ∈ s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ b. b ∈ s1 ↔ b = a ∨ b ∈ s0 ⊦ s1 = s0 ++ a

Rule argument

Additional information

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