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Proof step 9 from proof ext-setequal (previous step)

(∀ a. a ∈ x → a ∈ y) ∧ (∀ a. a ∈ y → a ∈ x) ⊦ ∀ a. a ∈ x ↔ a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ a. a ∈ x → a ∈ y, ∀ a. a ∈ y → a ∈ x ⊦ a ∈ x ↔ a ∈ y

Rule argument

Additional information