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Proof step 2 from proof prefix-subseteq-dom-nodes-rev (previous step)

st0 ⊑ st1 ⊦ 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 3:
⊦ st0 ⊑ st ↔ dom st0 ⊆ dom st ∧ (∀ n. n ∈ st0 ∧ n ∈ st → st0[n].list ⊑ st[n].list ∧ st0[n].ladrlist ⊑ st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist)
Proof step 4:
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)

Rule argument

spec: ablock-store-inv name: blockstore-prefix seq: ⊦ st0 ⊑ st ↔ dom st0 ⊆ dom st ∧ (∀ n. n ∈ st0 ∧ n ∈ st → st0[n].list ⊑ st[n].list ∧ st0[n].ladrlist ⊑ st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist) subst: [st, st0] → [st1, st0] flags: <all paths>

Additional information

Used terms: st1, st0