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Proof step 6 from proof diff-eq-diff-eq-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: set-union, inst: node-store name: Difference seq: ⊦ adr ∈ as \ as0 ↔ adr ∈ as ∧ ¬ adr ∈ as0 subst: [as0, as, adr] → [as0, as2, adr] flags: <all paths>

Additional information

Used terms: as0, as2, adr