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Proof step 4 from proof size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information