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Proof step 18 from proof prefix-subseteq-dom-nodes-rev (previous step)

adr = @(m, m0, m1), adr ∈ st0, ∀ n. n ∈ st0 → n ∈ 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 ∈ st1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
@(m, m0, m1) ∈ 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 ∈ st0 → n ∈ st1 ⊦ @(m, m0, m1) ∈ st1

Proof step 18 from proof prefix-subseteq-dom-nodes-rev (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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