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Proof step 21 from proof prefix-subseteq-dom-nodes-rev (previous step)

@(m, m0, m1).lnum ∈ st0, @(@(m, m0, m1).pos, @(m, m0, m1).size) ∈ st0[@(m, m0, m1).lnum].ladrlist, ∀ 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 ⊦ @(m, m0, m1) ∈ st1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
@(m0, m1) ∈ st0[m].ladrlist, m ∈ 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

Rule argument

Additional information

Used simplifier rules: ⊦ @(m, m0, m1).lnum = m; (s)% ⊦ @(m, m0, m1).size = m1; (s)% ⊦ @(m, m0, m1).pos = m0; (s)