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Proof step 6 from proof size (previous step)

∀ a. a ∈ y → a ∈ x, ∀ a. a ∈ x → a ∈ y ⊦ # x = # y, a ∈ x ↔ a ∈ y, dups(x), dups(y)

User applied rule case distinction

And obtained the following new goals

Proof step 7:
a ∈ x, ∀ a. a ∈ x → a ∈ y, ∀ a. a ∈ y → a ∈ x ⊦ # x = # y, dups(y), dups(x), a ∈ y
Proof step 8:
a ∈ y, ∀ a. a ∈ x → a ∈ y, ∀ a. a ∈ y → a ∈ x ⊦ # x = # y, dups(y), dups(x), a ∈ x

Rule argument

pos: right 2

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