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Proof step 8 from proof prefix-at (previous step)

st0[m] = mkblock(x, lax, m2), m ∈ st0, @(m0, m1) ∈ lax, dom st0 ⊆ dom st1, m ∈ st1 → x ⊑ st1[m].list ∧ lax ⊑ st1[m].ladrlist ∧ # x = # lax ∧ # st1[m].list = # st1[m].ladrlist ⊦ x[pos(@(m0, m1), lax)] = st1[@(m, m0, m1)]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
st0[m] = mkblock(x, lax, m2), st1[m] = mkblock(x0, lax0, m3), @(m0, m1) ∈ lax, m ∈ st0, dom st0 ⊆ dom st1, m ∈ st1 → x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0 ⊦ 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] → [st1[m], x0, lax0, m5, m6, m7, boolvar0]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st1[m], x0, lax0, m5, m6, m7, boolvar0