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Proof step 4 from proof prefix-just-alloc (previous step)

st1[n0] = mkblock(x, lax, m), n0 ∈ st0, n0 ∈ st1, n ∈ st0, dom st1 ⊆ dom st0, ∀ n0. n0 ∈ st1[n, mkblock([], [], 0)] ∧ n0 ∈ st0 → st1[n, mkblock([], [], 0)][n0].list ⊑ st0[n0].list ∧ st1[n, mkblock([], [], 0)][n0].ladrlist ⊑ st0[n0].ladrlist ∧ # st1[n, mkblock([], [], 0)][n0].list = # st1[n, mkblock([], [], 0)][n0].ladrlist ∧ # st0[n0].list = # st0[n0].ladrlist ⊦ n ∈ st1, x ⊑ st0[n0].list ∧ lax ⊑ st0[n0].ladrlist ∧ # x = # lax ∧ # st0[n0].list = # st0[n0].ladrlist

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
st1[n0] = mkblock(x, lax, m), st0[n0] = mkblock(x0, lax0, m0), n ∈ st0, n0 ∈ st1, n0 ∈ st0, dom st1 ⊆ dom st0, ∀ n0. n0 ∈ st1[n, mkblock([], [], 0)] ∧ n0 ∈ st0 → st1[n, mkblock([], [], 0)][n0].list ⊑ st0[n0].list ∧ st1[n, mkblock([], [], 0)][n0].ladrlist ⊑ st0[n0].ladrlist ∧ # st1[n, mkblock([], [], 0)][n0].list = # st1[n, mkblock([], [], 0)][n0].ladrlist ∧ # st0[n0].list = # st0[n0].ladrlist ⊦ n ∈ st1, x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0

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[n0], x0, lax0, m2, m3, m4, boolvar0]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st0[n0], x0, lax0, m2, m3, m4, boolvar0