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Proof step 11 from proof diff-eq-diff-eq-union (previous step)

as0 ⊆ as2, as1 ⊆ as2, as0 ⊆ as1, adr ∈ as2 \ as1 ∨ adr ∈ as1 \ as0 ⊦ adr ∈ as2 ∧ ¬ adr ∈ 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] → [as1, as2, adr] flags: <all paths>

Additional information

Used terms: as1, as2, adr