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Proof step 14 from proof prefix-rem-alloc (previous step)

(st0 \ (ns ++ n))[n0] = mkblock(x, lax, m), (st[n, ablock] \ (ns -- n))[n0] = mkblock(x0, lax0, m0), n1 ∈ st0, ∀ 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. n ∈ dom st0 \ ns → n ∈ dom st \ ns ⊦ n1 = n, n1 ∈ ns, n ∈ st, n1 ∈ (dom st ++ n) \ (ns -- n)

User applied rule all left

And obtained the following new goals

Proof step 15:
(st[n, ablock] \ (ns -- n))[n0] = mkblock(x0, lax0, m0), (st0 \ (ns ++ n))[n0] = mkblock(x, lax, m), n1 ∈ st0, n1 ∈ dom st0 \ ns → n1 ∈ 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 ⊦ n1 = n, n1 ∈ (dom st ++ n) \ (ns -- n), n ∈ st, n1 ∈ ns

Rule argument

pos: left 5 terms: n1 abort: F discard: T computed: F

Additional information

Used terms: n1