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Proof step 1 from proof ext (conclusion)
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
∀ a. a ∈ s1 ↔ a ∈ s2 ⊦ s1 = s2
Rule argument
Additional information