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Proof step 3 from proof subseteq-dom-nodes-alloc-restrict-empty (previous step)

as ⊆ dom nodes(st) ⊦ #(as | n) = 0, as | n = ∅, n ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ as = as0 ↔ (∀ adr. adr ∈ as ↔ adr ∈ as0)
Proof step 5:
as ⊆ dom nodes(st) ⊦ #(as | n) = 0, n ∈ st, ∀ adr. adr ∈ as | n ↔ adr ∈ ∅

Rule argument

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

Additional information

Used terms: ∅, as | n