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Proof step 1 from proof eq-04 (conclusion)

⊦ s ++ b ++ a = s ++ a ++ b ↔ true

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ a0 ∈ s ++ b ++ a ↔ a0 ∈ s ++ a ++ b

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);