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Proof step 10 from proof ext-setequal (previous step)

∀ a. a ∈ x → a ∈ y, ∀ a. a ∈ y → a ∈ x ⊦ a ∈ x ↔ a ∈ y

User applied rule case distinction

And obtained the following new goals

Proof step 11:
a ∈ x, ∀ a. a ∈ x → a ∈ y, ∀ a. a ∈ y → a ∈ x ⊦ a ∈ y
Proof step 12:
a ∈ y, ∀ a. a ∈ x → a ∈ y, ∀ a. a ∈ y → a ∈ x ⊦ a ∈ x

Rule argument

pos: right 1

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