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Proof step 6 from proof prefix-set-flags (previous step)

st[n] = mkblock(x, lax, m), n ∈ st, n1 ∈ st0, n1 ∈ st[n, mkblock(x, lax, n0)], 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 ⊦ st0[n1].list ⊑ st[n, mkblock(x, lax, n0)][n1].list ∧ st0[n1].ladrlist ⊑ st[n, mkblock(x, lax, n0)][n1].ladrlist ∧ # st0[n1].list = # st0[n1].ladrlist ∧ # st[n, mkblock(x, lax, n0)][n1].list = # st[n, mkblock(x, lax, n0)][n1].ladrlist

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
st[n] = mkblock(x, lax, m), st0[n1] = mkblock(x0, lax0, m0), n1 ∈ st[n, mkblock(x, lax, 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 ⊦ x0 ⊑ st[n, mkblock(x, lax, n0)][n1].list ∧ lax0 ⊑ st[n, mkblock(x, lax, n0)][n1].ladrlist ∧ # x0 = # lax0 ∧ # st[n, mkblock(x, lax, n0)][n1].list = # st[n, mkblock(x, lax, n0)][n1].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, lax, m] → [st0[n1], x0, lax0, m0]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st0[n1], x0, lax0, m0