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Proof step 8 from proof ext-setequal (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 9:
(∀ a. a ∈ x → a ∈ y) ∧ (∀ a. a ∈ y → a ∈ x) ⊦ ∀ a. a ∈ x ↔ a ∈ y
Proof step 13:
∀ a. a ∈ x ↔ a ∈ y ⊦ (∀ a. a ∈ x → a ∈ y) ∧ (∀ a. a ∈ y → a ∈ x)

Rule argument

pos: right 1

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