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Proof step 6 from proof prefix-subseteq-dom-nodes-rev (previous step)

dom st0 ⊆ dom st1, ∀ 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 ⊦ (∀ adr. adr ∈ dom nodes(st1) → adr ∈ dom nodes(st0)) ↔ dom nodes(st1) = dom nodes(st0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
dom st0 ⊆ dom st1, ∀ 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 ⊦ (∀ adr. adr ∈ st1 → adr ∈ st0) ↔ dom nodes(st1) = dom nodes(st0)

Rule argument

Additional information

Used simplifier rules: in-nodes: ⊦ adr ∈ nodes(st) ↔ adr ∈ st;% ⊦ adr ∈ dom ast ↔ adr ∈ ast; (s)