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Proof step 6 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 7:
a ∈ y, ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y, dups(x), dups(y), a ∈ x
Proof step 8:
a ∈ x, ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y, dups(x), dups(y), a ∈ y

Rule argument

pos: right 2

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