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Proof step 5 from proof dups-eqsize-ext (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 6:
a ∈ y, ∀ a. a ∈ x ↔ a ∈ y ⊦ a ∈ x, # x = # y, dups(y), dups(x)
Proof step 7:
a ∈ x, ∀ a. a ∈ x ↔ a ∈ y ⊦ a ∈ y, # x = # y, dups(y), dups(x)

Rule argument

pos: right 2

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