properties of Ackermann function

properties of Ackermann function

In this entry, we derive some basic

properties

of the

Ackermann function

A

⁢

(

x

,

y

)

, defined by double recursion, as follows:

A

⁢

(

0

,

y

)

=

y

+

1

,

A

⁢

(

x

+

1

,

0

)

=

A

⁢

(

x

,

1

)

,

A

⁢

(

x

+

1

,

y

+

1

)

=

A

⁢

(

x

,

A

⁢

(

x

+

1

,

y

)

)

.

These properties will be useful in proving that

A

is not

primitive recursive

.

1.

A

is total (

dom

⁡

(

A

)

=

ℕ

2

).

2.

A

⁢

(

1

,

y

)

=

y

+

2

.

3.

A

⁢

(

2

,

y

)

=

2

⁢

y

+

3

.

4.

y

<

A

⁢

(

x

,

y

)

.

5.

A

⁢

(

x

,

y

)

<

A

⁢

(

x

,

y

+

1

)

.

6.

A

⁢

(

x

,

y

+

1

)

≤

A

⁢

(

x

+

1

,

y

)

.

7.

A

⁢

(

x

,

y

)

<

A

⁢

(

x

+

1

,

y

)

.

8.

A

⁢

(

r

,

A

⁢

(

s

,

y

)

)

<

A

⁢

(

r

+

s

+

2

,

y

)

9.

For any

r

,

s

,

A

⁢

(

r

,

y

)

+

A

⁢

(

s

,

y

)

<

A

⁢

(

t

,

y

)

for some

t

not depending on

y

.

Most of the proofs are done by

induction

.

Proof.

1.

Induct on

x

. First,

A

⁢

(

0

,

y

)

=

y

+

1

is

well-defined

, so

(

0

,

y

)

∈

dom

⁡

(

A

)

for all

y

. Next, suppose that for a given

x

,

(

x

,

y

)

∈

dom

⁡

(

A

)

for all

y

. We want to show that

(

x

+

1

,

y

)

∈

dom

⁡

(

A

)

for all

y

. To do this, induct on

y

. First,

(

x

+

1

,

0

)

∈

dom

⁡

(

A

)

, since

A

⁢

(

x

+

1

,

0

)

=

A

⁢

(

x

,

1

)

is well-defined. Next, assume that

(

x

+

1

,

y

)

∈

dom

⁡

(

A

)

. Then

A

⁢

(

x

,

A

⁢

(

x

+

1

,

y

)

)

=

A

⁢

(

x

+

1

,

y

+

1

)

is well-defined. so

(

x

+

1

,

y

+

1

)

∈

dom

⁡

(

A

)

as well.

2.

Induct on

y

. First,

A

⁢

(

1

,

0

)

=

A

⁢

(

0

,

1

)

=

2

. Next, assume

A

⁢

(

1

,

y

)

=

y

+

2

. Then

A

⁢

(

1

,

y

+

1

)

=

A

⁢

(

0

,

y

+

2

)

=

y

+

3

=

(

y

+

1

)

+

2

.

3.

Induct on

y

. First,

A

⁢

(

2

,

0

)

=

A

⁢

(

1

,

1

)

=

1

+

2

=

3

. Next, assume

A

⁢

(

2

,

y

)

=

2

⁢

y

+

3

. Then

A

⁢

(

2

,

y

+

1

)

=

A

⁢

(

1

,

A

⁢

(

2

,

y

)

)

=

A

⁢

(

2

,

y

)

+

2

=

(

2

⁢

y

+

3

)

+

2

=

2

⁢

(

y

+

1

)

+

3

.

4.

Induct on

x

. First,

y

<

y

+

1

=

A

⁢

(

0

,

y

)

. Next, assume

y

<

A

⁢

(

x

,

y

)

, where

x

>

0

. Then

y

+

1

≤

A

⁢

(

x

,

y

)

<

A

⁢

(

x

-

1

,

A

⁢

(

x

,

y

)

)

=

A

⁢

(

x

,

y

+

1

)

.

5.

Induct on

x

. First,

A

⁢

(

0

,

y

)

=

y

+

1

<

y

+

2

=

A

⁢

(

0

,

y

+

1

)

. Next, assume that

A

⁢

(

x

,

y

)

<

A

⁢

(

x

,

y

+

1

)

. Then

A

⁢

(

x

+

1

,

y

)

<

A

⁢

(

x

,

A

⁢

(

x

+

1

,

y

)

)

=

A

⁢

(

x

+

1

,

y

+

1

)

.

6.

Induct on

y

. First,

A

⁢

(

x

,

1

)

=

A

⁢

(

x

+

1

,

0

)

. Next, assume that

A

⁢

(

x

,

y

+

1

)

≤

A

⁢

(

x

+

1

,

y

)

. Then

A

⁢

(

x

,

y

+

2

)

≤

A

⁢

(

x

,

A

⁢

(

x

,

y

+

1

)

)

≤

A

⁢

(

x

,

A

⁢

(

x

+

1

,

y

)

)

=

A

⁢

(

x

+

1

,

y

+

1

)

.

7.

Induct on

x

. First,

A

⁢

(

0

,

y

)

=

y

+

1

<

y

+

2

=

A

⁢

(

1

,

y

)

. Next, assume that

A

⁢

(

x

,

y

)

<

A

⁢

(

x

+

1

,

y

)

. There are two cases:

y

=

0

. Then

A

⁢

(

x

+

1

,

0

)

=

A

⁢

(

x

,

1

)

<

A

⁢

(

x

+

1

,

1

)

. Otherwise,

y

=

z

+

1

, so that

A

⁢

(

x

+

1

,

y

)

=

A

⁢

(

x

+

1

,

z

+

1

)

=

A

⁢

(

x

,

A

⁢

(

x

+

1

,

z

)

)

≤

A

⁢

(

x

,

A

⁢

(

x

,

z

+

1

)

)

=

A

⁢

(

x

,

A

⁢

(

x

,

y

)

)

<

A

⁢

(

x

,

A

⁢

(

x

+

1

,

y

)

)

=

A

⁢

(

x

+

1

,

y

+

1

)

.

8.

A

⁢

(

r

,

A

⁢

(

s

,

y

)

)

<

A

⁢

(

r

+

s

,

A

⁢

(

s

,

y

)

)

<

A

⁢

(

r

+

s

,

A

⁢

(

r

+

s

+

1

,

y

)

)

=

A

⁢

(

r

+

s

+

1

,

y

+

1

)

≤

A

⁢

(

r

+

s

+

2

,

y

)

.

9.

Let

z

=

max

⁡

{

r

,

s

}

. Then

A

⁢

(

r

,

y

)

+

A

⁢

(

s

,

y

)

≤

2

⁢

A

⁢

(

z

,

y

)

<

2

⁢

A

⁢

(

z

,

y

)

+

3

=

A

⁢

(

2

,

A

⁢

(

z

,

y

)

)

<

A

⁢

(

4

+

z

,

y

)

. The proof is completed by setting

t

=

4

+

z

.

∎

With respect to the

recursive

property of

A

, we see that

A

is recursive, since, by Church’s Thesis,

A

can be effectively computed (in fact, a program can be easily written to compute

A

⁢

(

x

,

y

)

). We also have the following:

Proposition 1

.

Define

A

n

:

N

→

N

by

A

n

⁢

(

m

)

=

A

⁢

(

n

,

m

)

. Then

A

n

is primitive recursive for each

n

.

Proof.

A

0

⁢

(

m

)

=

m

+

1

=

s

⁢

(

m

)

, so is primitive recursive. Now, assume

A

n

is primitive recursive, then

A

n

+

1

⁢

(

0

)

=

A

⁢

(

n

+

1

,

0

)

=

A

⁢

(

n

,

1

)

=

A

n

⁢

(

1

)

=

k

, and

A

n

+

1

⁢

(

m

+

1

)

=

A

⁢

(

n

+

1

,

m

+

1

)

=

A

⁢

(

n

,

A

⁢

(

n

+

1

,

m

)

)

=

A

n

⁢

(

A

⁢

(

n

+

1

,

m

)

)

=

A

n

⁢

(

A

n

+

1

⁢

(

m

)

)

, so that

A

n

+

1

is defined by

primitive recursion

via the

constant function

const

k

, and

A

n

, which is primitive recursive by the

induction hypothesis

. Therefore

A

n

+

1

is primitive recursive also.
∎

The most important fact about

A

concerning recursiveness is that

A

is not primitive recursive. Due to the length of its proof, it is demonstrated in this entry (

http://planetmath.org/AckermannFunctionIsNotPrimitiveRecursive

).

Title

properties of Ackermann function

Canonical

name

PropertiesOfAckermannFunction

Date of creation

2013-03-22 19:07:25

Last modified on

2013-03-22 19:07:25

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

9

Author

CWoo (3771)

Entry type

Result

Classification

msc 03D75

Generated on Thu Feb 8 20:06:26 2018 by

LaTeXML