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Proof step 12 from proof inter-union-diff-subseteq (previous step)

adr ∈ as3, as1 ∩ as3 ⊆ as, ∀ adr. adr ∈ as0 ∩ as3 → adr ∈ as, adr ∈ as0 ∨ adr ∈ as1 \ as2 ⊦ #(as0 ∪ (as1 \ as2)) = 0, # as3 = 0, as3 = ∅, as0 ∪ (as1 \ as2) = ∅, adr ∈ as

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] → [as2, as1, adr] flags: <all paths>

Additional information

Used terms: as2, as1, adr