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Proof step 5 from proof prefix-at (previous step)

m ∈ st0, @(m0, m1) ∈ st0[m].ladrlist, dom st0 ⊆ dom 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 ⊦ st0[m].list[pos(@(m0, m1), st0[m].ladrlist)] = st1[@(m, m0, m1)]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
st0[m] = mkblock(x, lax, m2), @(m0, m1) ∈ lax, m ∈ st0, dom st0 ⊆ dom 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 ⊦ x[pos(@(m0, m1), lax)] = st1[@(m, m0, m1)]

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, lax, m, m0, m1, boolvar] → [st0[m], x, lax, m2, m3, m4, boolvar]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st0[m], x, lax, m2, m3, m4, boolvar