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Proof step 12 from proof prefix-append-in (previous step)

# lax0 = # y, st[n1] = mkblock(x1, lax2, m1), st[n] = mkblock(x, lax, m0), st0[n1] = mkblock(x0, lax1, m), n1 ∈ st[n, mkblock(x + y, lax + lax0, n0)], n1 ∈ st0, 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 ⊦ n1 = n, x0 ⊑ x1 ∧ lax1 ⊑ lax2 ∧ # x0 = # lax1 ∧ # x1 = # lax2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# lax0 = # y, st0[n1] = mkblock(x0, lax1, m), st[n] = mkblock(x, lax, m0), st[n1] = mkblock(x1, lax2, m1), n ∈ st, n1 ∈ st0, n1 ∈ 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 ⊦ n1 = n, x0 ⊑ x1 ∧ lax1 ⊑ lax2 ∧ # x0 = # lax1 ∧ # x1 = # lax2

Rule argument

Additional information

Used simplifier rules: ⊦ n ≠ n0 → (n ∈ st[n0, ablock] ↔ n ∈ st); (s)