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Proof step 6 from proof diff-empty-subseteq (previous step)

⊦ (∀ adr. ¬ adr ∈ as0 \ as1) ↔ (∀ adr. adr ∈ as0 → adr ∈ as1)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information