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Proof step 1 from proof dups-eqsize-02 (conclusion)

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

System applied rule normalize

And obtained the following new goals

Proof step 2:
# x = # y, ∀ a. a ∈ x → a ∈ y ⊦ dups(x), ¬ dups(y) ∧ (∀ a. a ∈ y → a ∈ x)

Rule argument

Additional information