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Proof step 14 from proof ext-setequal (previous step)

∀ a. a ∈ x ↔ a ∈ y ⊦ (a ∈ x → a ∈ y) ∧ (a0 ∈ y → a0 ∈ x)

User applied rule case distinction

And obtained the following new goals

Proof step 15:
∀ a. a ∈ x ↔ a ∈ y ⊦ a ∈ x → a ∈ y
Proof step 17:
∀ a. a ∈ x ↔ a ∈ y ⊦ a0 ∈ y → a0 ∈ x

Rule argument

pos: right 1

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