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Proof step 3 from proof prefix-rem-alloc (previous step)

dom st0 \ ns ⊆ dom st \ ns, ∀ n. n ∈ st0 \ ns ∧ n ∈ st \ ns → (st0 \ ns)[n].list ⊑ (st \ ns)[n].list ∧ (st0 \ ns)[n].ladrlist ⊑ (st \ ns)[n].ladrlist ∧ # (st0 \ ns)[n].list = # (st0 \ ns)[n].ladrlist ∧ # (st \ ns)[n].list = # (st \ ns)[n].ladrlist ⊦ n ∈ st, st0 \ (ns ++ n) ⊑ st[n, ablock] \ (ns -- n)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 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 5:
dom st0 \ ns ⊆ dom st \ ns, ∀ n. n ∈ st0 \ ns ∧ n ∈ st \ ns → (st0 \ ns)[n].list ⊑ (st \ ns)[n].list ∧ (st0 \ ns)[n].ladrlist ⊑ (st \ ns)[n].ladrlist ∧ # (st0 \ ns)[n].list = # (st0 \ ns)[n].ladrlist ∧ # (st \ ns)[n].list = # (st \ ns)[n].ladrlist ⊦ n ∈ st, dom(st0 \ (ns ++ n)) ⊆ dom(st[n, ablock] \ (ns -- n)) ∧ (∀ n0. n0 ∈ st0 \ (ns ++ n) ∧ n0 ∈ st[n, ablock] \ (ns -- n) → (st0 \ (ns ++ n))[n0].list ⊑ (st[n, ablock] \ (ns -- n))[n0].list ∧ (st0 \ (ns ++ n))[n0].ladrlist ⊑ (st[n, ablock] \ (ns -- n))[n0].ladrlist ∧ # (st0 \ (ns ++ n))[n0].list = # (st0 \ (ns ++ n))[n0].ladrlist ∧ # (st[n, ablock] \ (ns -- n))[n0].list = # (st[n, ablock] \ (ns -- n))[n0].ladrlist)

Rule argument

spec: <toplevel> 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] → [st[n, ablock] \ (ns -- n), st0 \ (ns ++ n)] flags: <all paths>

Additional information

Used terms: st[n, ablock] \ (ns -- n), st0 \ (ns ++ n)