Ackermann function is not primitive recursive

Ackermann function is not primitive recursive

In this entry, we show that the

Ackermann function

A

⁢

(

x

,

y

)

, given by

A

⁢

(

0

,

y

)

=

y

+

1

,

A

⁢

(

x

+

1

,

0

)

=

A

⁢

(

x

,

1

)

,

A

⁢

(

x

+

1

,

y

+

1

)

=

A

⁢

(

x

,

A

⁢

(

x

+

1

,

y

)

)

is not

primitive recursive

. We will utilize the

properties

of

A

listed in this entry (

http://planetmath.org/PropertiesOfAckermannFunction

).

The key to showing that

A

is not primitive recursive, is to find a properties shared by all

primitive recursive functions

, but not by

A

. One such property is in showing that

A

in some way “grows” faster than any primitive recursive function. This is formalized by the notion of “

majorization

”, which is explained here (

http://planetmath.org/SuperexponentiationIsNotElementary

).

Proposition 1

.

Let

A

be the set of all

functions

majorized by

A

. Then

P

⁢

R

⊆

A

.

Proof.

We break this up into three parts: show all initial functions are in

𝒜

, show

𝒜

is

closed under

functional

composition

, and show

𝒜

is closed under

primitive recursion

. The proof is completed by realizing that

𝒫

⁢

ℛ

is the smallest set satisfying the three conditions.

In the proofs below,

𝒙

denotes some tuple of non-negative

integers

(

x

1

,

…

,

x

n

)

for some

n

, and

x

=

max

⁡

{

x

1

,

…

,

x

n

}

. Likewise for

𝒚

and

y

.

1.

We show that the

zero function

, the successor function, and the projection functions are in

𝒜

.

–

z

⁢

(

n

)

=

0

<

n

+

1

=

A

⁢

(

0

,

n

)

, so

z

∈

𝒜

.

–

s

⁢

(

n

)

=

n

+

1

<

n

+

2

=

A

⁢

(

1

,

n

)

, so

s

∈

𝒜

.

–

p

m

k

⁢

(

x

1

,

…

,

x

k

)

=

x

m

≤

x

<

x

+

1

=

A

⁢

(

0

,

x

)

, so

p

m

k

∈

𝒜

.

2.

Next, suppose

g

1

,

…

,

g

m

are

k

-ary, and

h

is

m

-ary, and that each

g

i

, and

h

are in

𝒜

. This means that

g

i

⁢

(

𝒙

)

<

A

⁢

(

r

i

,

x

)

, and

h

⁢

(

𝒚

)

<

A

⁢

(

s

,

y

)

. Let

f

=

h

⁢

(

g

1

,

…

,

g

m

)

,

and

g

j

⁢

(

𝒙

)

=

max

⁡

{

g

i

⁢

(

𝒙

)

∣

i

=

1

,

…

,

m

}

.

Then

f

⁢

(

𝒙

)

<

A

⁢

(

s

,

g

j

⁢

(

𝒙

)

)

<

A

⁢

(

s

,

A

⁢

(

r

j

,

x

)

)

<

A

⁢

(

s

+

r

j

+

2

,

x

)

, showing that

f

∈

𝒜

.

3.

Finally, suppose

g

is

k

-ary and

h

is

(

k

+

2

)

-ary, and that

g

,

h

∈

𝒜

. This means that

g

⁢

(

𝒙

)

<

A

⁢

(

r

,

x

)

and

h

⁢

(

𝒚

)

<

A

⁢

(

s

,

y

)

. We want to show that

f

, defined by primitive recursion via functions

g

and

h

, is in

𝒜

.

We first prove the following claim:

f

⁢

(

𝒙

,

n

)

<

A

⁢

(

q

,

n

+

x

)

,

for some

⁢

q

⁢

not depending on

⁢

x

⁢

and

⁢

n

.

Pick

q

=

1

+

max

⁡

{

r

,

s

}

, and induct on

n

. First,

f

⁢

(

𝒙

,

0

)

=

g

⁢

(

𝒙

)

<

A

⁢

(

r

,

x

)

<

A

⁢

(

q

,

x

)

. Next, suppose

f

⁢

(

𝒙

,

n

)

<

A

⁢

(

q

,

n

+

x

)

. Then

f

⁢

(

𝒙

,

n

+

1

)

=

h

⁢

(

𝒙

,

n

,

f

⁢

(

𝒙

,

n

)

)

<

A

⁢

(

s

,

z

)

, where

z

=

max

⁡

{

x

,

n

,

f

⁢

(

𝒙

,

n

)

}

. By the

induction hypothesis

, together with the fact that

max

⁡

{

x

,

n

}

≤

n

+

x

<

A

⁢

(

q

,

n

+

x

)

, we see that

z

<

A

⁢

(

q

,

n

+

x

)

. Thus,

f

⁢

(

𝒙

,

n

+

1

)

<

A

⁢

(

s

,

z

)

<

A

⁢

(

s

,

A

⁢

(

q

,

n

+

x

)

)

≤

A

⁢

(

q

-

1

,

A

⁢

(

q

,

n

+

x

)

)

=

A

⁢

(

q

,

n

+

1

+

x

)

, proving the claim.

To finish the proof, let

z

=

max

⁡

{

x

,

y

}

. Then, by the claim,

f

⁢

(

𝒙

,

y

)

<

A

⁢

(

q

,

x

+

y

)

≤

A

⁢

(

q

,

2

⁢

z

)

<

A

⁢

(

q

,

2

⁢

z

+

3

)

=

A

⁢

(

q

,

A

⁢

(

2

,

z

)

)

=

A

⁢

(

q

+

4

,

z

)

, showing that

f

∈

𝒜

.

Since

𝒫

⁢

ℛ

is by

definition

the smallest set containing the initial functions, and closed under composition and primitive recursion,

𝒫

⁢

ℛ

⊆

𝒜

.
∎

As a

corollary

, we have

Corollary 1

.

The Ackermann function

A

is not primitive recursive.

Proof.

Otherwise,

A

∈

𝒜

, which is impossible. ∎

Title

Ackermann function is not primitive recursive

Canonical

name

AckermannFunctionIsNotPrimitiveRecursive

Date of creation

2013-03-22 19:07:29

Last modified on

2013-03-22 19:07:29

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

11

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 03D75

Related topic

AckermannFunctionIsTotalRecursive

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