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Proof step 7 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 8:
⊦ (∀ a. a ∈ x → a ∈ y) ∧ (∀ a. a ∈ y → a ∈ x) ↔ (∀ a. a ∈ x ↔ a ∈ y)

Rule argument

Additional information