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Proof step 18 from proof prefix-rem-alloc (previous step)

(st[n, ablock] \ (ns -- n))[n0] = mkblock(x0, lax0, m0), st0[n0] = mkblock(x, lax, m), 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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ ¬ n ∈ ns → (st \ ns)[n] = st[n]
Proof step 20:
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

Rule argument

spec: store+set, inst: ablock-store-nat-store name: Diff-get seq: ⊦ ¬ n ∈ ns → (st \ ns)[n] = st[n] subst: [n, ns, st] → [n0, ns -- n, st[n, ablock]] flags: <all paths>

Additional information

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