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Proof step 4 from proof dups-eqsize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ x, ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y, dups(y), dups(x), a ∈ y

Rule argument

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