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Proof step 8 from proof diff-subseteq-subseteq (previous step)

as1 ⊆ as2 ⊦ (∀ adr. adr ∈ as0 \ as1 → adr ∈ as2) ↔ (∀ adr. adr ∈ as0 → adr ∈ as2)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information