310 Miscellaneous Exercises
(b) Show that the values m(i) cannot be obtained effectively.
∗∗
(c) Show that it is impossible to obtain a machine for A running in
time f
n−k
(2) effectively from k.
106. Let P be a nontrivial property of the linear-time sets that is true for all
regular sets. Show that P is undecidable.
∗HS
107. This is a general isomorphism theorem that has both the Rogers isomor-
phism theorem (Theorem 34.3, Miscellaneous Exercise 108) and the My-
hill isomorphism theorem (Miscellaneous Exercise 109) as special cases.
Let ◦ denote relational composition and
−1
the reverse operator on bi-
nary relations on ω.Thatis,forR, S ⊆ ω × ω, define
R ◦ S
def
= {(u, w) |∃v (u, v) ∈ R, (v, w) ∈ S}
R
−1
def
= {(u, v) | (v, u) ∈ R}.
For a function f : ω → ω, define
graph f
def
= {(x, f (x)) | x ∈ ω}.
Let R be a binary relation on ω such that R ◦ R
−1
◦ R ⊆ R.Letf,g :
ω → ω be one-to-one total recursive functions such that graph f ⊆ R
and graph g ⊆ R
−1
. Show that there exists a one-to-one and onto total
recursive function h : ω → ω such that graph h ⊆ R.
H
108. Here we use Miscellaneous Exercise 107 to give an alternative proof of
the Rogers isomorphism theorem (Theorem 34.3). Assuming there exist
σ, τ : ω → ω such that for all i, ϕ
i
= ψ
σ(i)
and ψ
i
= ϕ
τ (i)
, show that
there exists a one-to-one and onto total recursive function ρ : ω → ω
such that for all i, ϕ
i
= ψ
ρ(i)
.
H
109. The Cantor–Schr¨oder–Bernstein theorem of set theory says that if there
is a one-to-one function A → B and a one-to-one function B → A,then
A and B are of the same cardinality. Here is an effective version due
to Myhill [90] (see [104]) known as the Myhill isomorphism theorem.
Show that any two sets that are reducible to each other via one-to-one
reductions are recursively isomorphic. In other words, let A, B ⊆ ω and
let f,g : ω → ω be one-to-one total recursive functions such that for all
x ∈ ω, x ∈ A ⇔ f(x) ∈ B and x ∈ B ⇔ g(x) ∈ A. Show that there
exists a one-to-one and onto total recursive function h : ω → ω such
that for all x ∈ ω, x ∈ A ⇔ h(x) ∈ B.