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Proof step 8 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) ↔ 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

User applied rule case distinction

And obtained the following new goals

Proof step 9:
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)

Rule argument

pos: right 1

Additional information

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