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

# lax0 = # y, st[n] = mkblock(x, lax, m0), n ∈ st, dom st0 ⊆ dom st, ∀ n. n ∈ st0 ∧ n ∈ st → st0[n].list ⊑ st[n].list ∧ st0[n].ladrlist ⊑ st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ⊦ dom st0 ⊆ dom st[n, mkblock(x + y, lax + lax0, n0)] ∧ (∀ n1. n1 ∈ st0 ∧ n1 ∈ st[n, mkblock(x + y, lax + lax0, n0)] → st0[n1].list ⊑ st[n, mkblock(x + y, lax + lax0, n0)][n1].list ∧ st0[n1].ladrlist ⊑ st[n, mkblock(x + y, lax + lax0, n0)][n1].ladrlist ∧ # st0[n1].list = # st0[n1].ladrlist ∧ # st[n, mkblock(x + y, lax + lax0, n0)][n1].list = # st[n, mkblock(x + y, lax + lax0, n0)][n1].ladrlist)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# lax0 = # y, st[n] = mkblock(x, lax, m0), n ∈ st, n1 ∈ st0, n1 ∈ st[n, mkblock(x + y, lax + lax0, n0)], dom st0 ⊆ dom st, ∀ n. n ∈ st0 ∧ n ∈ st → st0[n].list ⊑ st[n].list ∧ st0[n].ladrlist ⊑ st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ⊦ st0[n1].list ⊑ st[n, mkblock(x + y, lax + lax0, n0)][n1].list ∧ st0[n1].ladrlist ⊑ st[n, mkblock(x + y, lax + lax0, n0)][n1].ladrlist ∧ # st0[n1].list = # st0[n1].ladrlist ∧ # st[n, mkblock(x + y, lax + lax0, n0)][n1].list = # st[n, mkblock(x + y, lax + lax0, n0)][n1].ladrlist

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ st → dom st ++ n = dom st; (s)% ⊦ dom st[n, ablock] = dom st ++ n; (s)