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Proof step 1 from proof size (conclusion)

⊦ ¬ dups(x) ∧ ¬ dups(y) ∧ x ⊆ y ∧ y ⊆ x → (# x = # y ↔ true)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
y ⊆ x, x ⊆ y ⊦ # x = # y, dups(x), dups(y)

Rule argument

Additional information

Used simplifier rules: subset-trans: ⊦ x ⊆ y ∧ y ⊆ z → x ⊆ z;%subset-ref: ⊦ x ⊆ x;