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Proof step 2 from proof size (previous step)

x ⊆ y, y ⊆ x ⊦ # x = # y ↔ true, dups(y), dups(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a. a ∈ y → a ∈ x, ∀ a. a ∈ x → a ∈ y ⊦ # x = # y, dups(x), dups(y)

Rule argument

Additional information

Used simplifier rules: subset: ⊦ x ⊆ y ↔ (∀ a. a ∈ x → a ∈ y);