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Proof step 3 from proof remove-refsize0-subset-refed (previous step)

as ⊆ dom nodes(st), ∀ adr. adr ∈ as → adr.size ≠ 0, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ nodes(st) | as ⊆ nodes(remove-refsize0(st))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ast ⊆ ast0 ↔ (∀ adr. adr ∈ ast → adr ∈ ast0 ∧ ast[adr] = ast0[adr])
Proof step 5:
as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size, ∀ adr. adr ∈ as → adr.size ≠ 0 ⊦ ∀ adr. adr ∈ nodes(st) | as → adr ∈ nodes(remove-refsize0(st)) ∧ (nodes(st) | as)[adr] = nodes(remove-refsize0(st))[adr]

Rule argument

spec: store, inst: address-store name: Subset seq: ⊦ ast ⊆ ast0 ↔ (∀ adr. adr ∈ ast → adr ∈ ast0 ∧ ast[adr] = ast0[adr]) subst: [ast0, ast] → [nodes(remove-refsize0(st)), nodes(st) | as] flags: <all paths>

Additional information

Used terms: nodes(remove-refsize0(st)), nodes(st) | as