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Proof step 5 from proof ext (previous step)

∀ a. a ∈ s1 ↔ a ∈ s2 ⊦ a ∈ s1 ↔ a ∈ s2

User applied rule case distinction

And obtained the following new goals

Proof step 6:
a ∈ s1, ∀ a. a ∈ s1 ↔ a ∈ s2 ⊦ a ∈ s2
Proof step 8:
a ∈ s2, ∀ a. a ∈ s1 ↔ a ∈ s2 ⊦ a ∈ s1

Rule argument

pos: right 1

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