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Proof step 4 from proof prefix-del-right (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ 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

Rule argument

Additional information

Used simplifier rules: ⊦ n ≠ n0 → (st -- n0)[n] = st[n]; (s)% ⊦ n ∈ st -- n0 ↔ n ≠ n0 ∧ n ∈ st; (s)% ⊦ dom(st -- n) = dom st -- n; (s)