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Proof step 10 from proof prefix-del-right (previous step)

n ∈ st, n ∈ st0, dom st ⊆ dom st0, ∀ n. n ∈ st ∧ n ∈ st0 → st[n].list ⊑ st0[n].list ∧ st[n].ladrlist ⊑ st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ⊦ m ∈ st, st[n].list ⊑ st0[n].list ∧ st[n].ladrlist ⊑ st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
st[n] = mkblock(x, lax, m0), n ∈ st0, n ∈ st, dom st ⊆ dom st0, ∀ n. n ∈ st ∧ n ∈ st0 → st[n].list ⊑ st0[n].list ∧ st[n].ladrlist ⊑ st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ⊦ m ∈ st, x ⊑ st0[n].list ∧ lax ⊑ st0[n].ladrlist ∧ # x = # lax ∧ # st0[n].list = # st0[n].ladrlist

Rule argument

spec: ablock name: elim seq: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m) subst: [ablock, x, m, m0, m1] → [st[n], x, m0, m1, m2]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st[n], x, m0, m1, m2