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Proof step 10 from proof diff-eq-diff-eq-union (previous step)

as0 ⊆ as1, as1 ⊆ as2, as0 ⊆ as2 ⊦ adr ∈ as2 ∧ ¬ adr ∈ as0 ↔ adr ∈ as2 \ as1 ∨ adr ∈ as1 \ as0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ as \ as0 ↔ adr ∈ as ∧ ¬ adr ∈ as0; (ws)