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

st0[n] = mkblock(x, lax, m), n ∈ st0, ∀ 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, ∀ n. n ∈ dom st0 → n ∈ dom st ⊦ n ∈ st, x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
st0[n] = mkblock(x, lax, m), n ∈ st0, ∀ 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, ∀ n0. n0 ∈ st0 → n0 ∈ st ⊦ n ∈ st, x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ dom st ↔ n ∈ st; (s)