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Proof step 20 from proof prefix-rem-alloc (previous step)

st0[n0] = mkblock(x, lax, m), st[n, ablock][n0] = mkblock(x0, lax0, m0), n0 ∈ st, n0 ∈ st0, dom st0 \ ns ⊆ dom st \ ns, ∀ n. n ∈ st0 \ ns ∧ n ∈ st \ ns → (st0 \ ns)[n].list ⊑ (st \ ns)[n].list ∧ (st0 \ ns)[n].ladrlist ⊑ (st \ ns)[n].ladrlist ∧ # (st0 \ ns)[n].list = # (st0 \ ns)[n].ladrlist ∧ # (st \ ns)[n].list = # (st \ ns)[n].ladrlist ⊦ n0 ∈ ns, n ∈ st, n0 ∈ ns -- n, x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
st0[n0] = mkblock(x, lax, m), st[n0] = mkblock(x0, lax0, m0), n0 ∈ st0, n0 ∈ st, dom st0 \ ns ⊆ dom st \ ns, ∀ n. n ∈ st0 \ ns ∧ n ∈ st \ ns → (st0 \ ns)[n].list ⊑ (st \ ns)[n].list ∧ (st0 \ ns)[n].ladrlist ⊑ (st \ ns)[n].ladrlist ∧ # (st0 \ ns)[n].list = # (st0 \ ns)[n].ladrlist ∧ # (st \ ns)[n].list = # (st \ ns)[n].ladrlist ⊦ n ∈ st, n0 ∈ ns, x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ ns -- n0 ↔ n ≠ n0 ∧ n ∈ ns; (s)% ⊦ n ∈ st ∧ ¬ n0 ∈ st → st[n0, ablock][n] = st[n]; (s)