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Proof step 4 from proof ext-rec (previous step)

∀ b. b ∈ s1 ↔ b = a ∨ b ∈ s0 ⊦ ∀ a0. a0 ∈ s1 ↔ a0 ∈ s0 ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ b. b ∈ s1 ↔ b = a ∨ b ∈ s0 ⊦ a0 ∈ s1 ↔ a0 = a ∨ a0 ∈ s0

Rule argument

Additional information

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