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Proof step 3 from proof diff-eq-diff-eq-union (previous step)

as0 ⊆ as1, as1 ⊆ as2, as0 ⊆ as2 ⊦ as2 \ as0 = (as2 \ as1) ∪ (as1 \ as0)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: set-basic, inst: node-store name: Extension seq: ⊦ as = as0 ↔ (∀ adr. adr ∈ as ↔ adr ∈ as0) subst: [as0, as] → [(as2 \ as1) ∪ (as1 \ as0), as2 \ as0] flags: <all paths>

Additional information

Used terms: (as2 \ as1) ∪ (as1 \ as0), as2 \ as0