90 J. Dassow
2.17 Example We consider the equivalence relation
1
defined by
1
= {(F, G) | arity(F )=arity(G),F((0, 0,...,0)) = G((0, 0,...,0))}.
It is easy to see that the equivalence classes of
1
are given by
K
n,a
= {F | arity(F )=n and F ((0, 0,...,0)) = a},
where n ∈ N and a ∈{0, 1}.
Obviously, F is a
1
-algebra (this statement holds for any equivalence relation). But there
are further
1
-algebras. Here we only present the subalgebra with the target set
Q = {F | F
p
(q)=0
|q|
for |p|≥1},
which is the set of all sequential functions where, for t ≥ 2, ϕ
t
F
is the constant k
0
.Itiseasy
to see that these functions form a subalgebra. Moreover, it is a
1
-subalgebra since, for any
Boolean function f, it contains a sequential function F with ϕ
1
F
= f.
Since by applications of ∇ and ∆, we can obtain functions of arbitrary arity, M is
1
-
complete if and only if [M] contains at least one F with F((0, 0,...,0)) = 0 and at least one
G with G((0, 0,...,0)) = 1.
If M only contains functions F with F ((0, 0,...,0)) = 0, then this holds for [M]too.
Therefore M contains a function G with G((0, 0,...,0)) = 1. Let G
be the sequential
function obtained from G by identification of all inputs.
First let us assume that G(1, 1,...,1) = 0. Then, ϕ
1
G
=non. Thusϕ
1
G
◦G
= id, i.e.,
(G
◦G
)(0) = 0. Thus M is
1
-complete.
We now assume that ϕ
1
G
(1, 1,...,1) = 1 and that the function ϕ
1
G
is not a constant.
Then there is a tuple (a
1
,a
2
,...,a
r
)withϕ
1
G
(a
1
,a
2
,...,a
r
)=0. For1≤ i ≤ r, we define
the sequential functions H
i
by
H
i
=
F
(id)
if a
i
=0
G
if a
i
=1
Then the sequential function H ∈F
1
defined by
H(p)=G(H
1
(p),H
2
(p),...,H
r
(p))
satisfies ϕ
1
H
(0) = 0. Thus M is also
1
-complete in this case.
Finally, let us assume that ϕ
1
G
is the (r-ary) constant k
1
.ThenM has to contain a
function F with F ((0, 0,...,0)) = 0 in order to be
1
-complete.
Thus we have the following decidable criterion for
1
-completeness: M is
1
-complete if
and only if M contains at least one function G where G((0, 0,...,0)) = 1 and ϕ
1
G
is not a
constant or M contains at least a function G where ϕ
1
G
is the constant k
1
and one F with
F ((0, 0,...,0)) = 0.
By this criterion (or already by the definition),
M
T
0
= {F | F ((0, 0,...,0)) = 0}
is an example for a
1
-maximal subalgebra.