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Proof step 5 from proof diff-eq-diff-eq-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information