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Proof step 1 from proof diff-empty-subseteq (conclusion)

⊦ as0 \ as1 = ∅ ↔ as0 ⊆ as1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ as = as0 ↔ (∀ adr. adr ∈ as ↔ adr ∈ as0)
Proof step 3:
⊦ (∀ adr. adr ∈ as0 \ as1 ↔ adr ∈ ∅) ↔ as0 ⊆ as1

Rule argument

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

Additional information

Used terms: ∅, as0 \ as1