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Proof step 10 from proof inter-union-diff-subseteq (previous step)

adr ∈ as3, adr ∈ as0 ∪ (as1 \ as2), as1 ∩ as3 ⊆ as, ∀ adr. adr ∈ as0 ∩ as3 → adr ∈ as ⊦ # as3 = 0, #(as0 ∪ (as1 \ as2)) = 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: Union seq: ⊦ adr ∈ as ∪ as0 ↔ adr ∈ as ∨ adr ∈ as0 subst: [as0, as, adr] → [as1 \ as2, as0, adr] flags: <all paths>

Additional information

Used terms: as1 \ as2, as0, adr