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Proof step 5 from proof prefix-del-right (previous step)

⊦ dom st ⊆ dom st0 -- m ∧ (∀ n. n ∈ st ∧ n ≠ m ∧ n ∈ st0 → st[n].list ⊑ st0[n].list ∧ st[n].ladrlist ⊑ st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist) ↔ st ⊑ st0 ∧ ¬ m ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 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 7:
⊦ dom st ⊆ dom st0 -- m ∧ (∀ n. n ∈ st ∧ n ≠ m ∧ n ∈ st0 → st[n].list ⊑ st0[n].list ∧ st[n].ladrlist ⊑ st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist) ↔ ( dom st ⊆ dom st0 ∧ (∀ n. n ∈ st ∧ n ∈ st0 → st[n].list ⊑ st0[n].list ∧ st[n].ladrlist ⊑ st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist)) ∧ ¬ m ∈ st

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] → [st0, st] flags: <all paths>

Additional information

Used terms: st0, st