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Proof step 2 from proof prefix-rem (previous step)

st0 \ ns ⊑ st \ ns ⊦ (st0 \ ns) -- n ⊑ (st \ ns) -- n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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 ⊦ (st0 \ ns) -- n ⊑ (st \ ns) -- n

Rule argument

Additional information

Used simplifier rules: blockstore-prefix: ⊦ 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);% ⊦ dom(st \ ns) = dom st \ ns; (s)