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Proof step 9 from proof prefix-rem-alloc (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ ns ⊆ ns0 ↔ (∀ n. n ∈ ns → n ∈ ns0)
Proof step 11:
(st0 \ (ns ++ n))[n0] = mkblock(x, lax, m), (st[n, ablock] \ (ns -- n))[n0] = mkblock(x0, lax0, m0), 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. n0 ∈ (dom st0 \ ns) -- n → n0 ∈ (dom st ++ n) \ (ns -- n)

Rule argument

spec: set-basic, inst: nat-set name: Subset seq: ⊦ ns ⊆ ns0 ↔ (∀ n. n ∈ ns → n ∈ ns0) subst: [ns0, ns] → [(dom st ++ n) \ (ns -- n), (dom st0 \ ns) -- n] flags: <all paths>

Additional information

Used terms: (dom st ++ n) \ (ns -- n), (dom st0 \ ns) -- n