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Proof step 5 from proof prefix-fwd-in (previous step)

n ∈ st0, ∀ n. n ∈ st0 ∧ n ∈ st1 → st0[n].list ⊑ st1[n].list ∧ st0[n].ladrlist ⊑ st1[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st1[n].list = # st1[n].ladrlist, ∀ n. n ∈ dom st0 → n ∈ dom st1 ⊦ n ∈ st1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n ∈ st0, ∀ n. n ∈ st0 ∧ n ∈ st1 → st0[n].list ⊑ st1[n].list ∧ st0[n].ladrlist ⊑ st1[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st1[n].list = # st1[n].ladrlist, ∀ n0. n0 ∈ st0 → n0 ∈ st1 ⊦ n ∈ st1

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ dom st ↔ n ∈ st; (s)