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Proof step 4 from proof prefix-subseteq-dom-nodes-rev (previous step)

dom st0 ⊆ dom st1, ∀ n. n ∈ st0 ∧ n ∈ st1 → st0[n].list ⊑ st1[n].list ∧ st0[n].ladrlist ⊑ st1[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st1[n].list = # st1[n].ladrlist ⊦ dom nodes(st1) ⊆ dom nodes(st0) ↔ dom nodes(st1) = dom nodes(st0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0)
Proof step 6:
dom st0 ⊆ dom st1, ∀ n. n ∈ st0 ∧ n ∈ st1 → st0[n].list ⊑ st1[n].list ∧ st0[n].ladrlist ⊑ st1[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st1[n].list = # st1[n].ladrlist ⊦ (∀ adr. adr ∈ dom nodes(st1) → adr ∈ dom nodes(st0)) ↔ dom nodes(st1) = dom nodes(st0)

Rule argument

spec: set-basic, inst: address-store name: Subset seq: ⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0) subst: [as0, as] → [dom nodes(st0), dom nodes(st1)] flags: <all paths>

Additional information

Used terms: dom nodes(st0), dom nodes(st1)