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Proof step 9 from proof prefix-del-right (previous step)

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, 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ dom st ↔ n ∈ st; (s)% ⊦ ns ⊆ ns0 -- n ↔ ns ⊆ ns0 ∧ ¬ n ∈ ns; (ws)