properties of superexponentiation

properties of superexponentiation

In this entry, we list some basic

properties

of the superexponetial

function

f

:

ℕ

2

→

ℕ

, defined recursively by

f

⁢

(

m

,

0

)

=

m

,

f

⁢

(

m

,

n

+

1

)

=

m

f

⁢

(

m

,

n

)

.

Furthermore, we set

f

⁢

(

0

,

n

)

:=

0

for all

n

.

Given

m

, the values of

f

are

m

,

m

m

,

m

m

m

,

⋯

,

m

.

⋅

m

,

⋯

,

where the evaluation of these values start from the top, for example:

3

3

3

=

3

81

.

Proposition 1

.

Suppose

x

,

y

,

z

∈

N

(including

0

), and for all except the first assertion,

x

>

1

.

1.

x

≤

f

⁢

(

x

,

y

)

.

2.

f

⁢

(

x

,

y

)

is increasing in both

arguments

.

3.

2

⁢

f

⁢

(

x

,

y

)

≤

f

⁢

(

x

,

y

+

1

)

.

4.

f

⁢

(

x

,

y

)

2

≤

f

⁢

(

x

,

y

+

1

)

.

5.

f

⁢

(

x

,

y

)

f

⁢

(

x

,

y

)

≤

f

⁢

(

x

,

y

+

2

)

6.

f

⁢

(

x

,

y

)

+

f

⁢

(

x

,

z

)

≤

f

⁢

(

x

,

1

+

max

⁡

{

y

,

z

}

)

.

7.

f

⁢

(

x

,

y

)

⋅

f

⁢

(

x

,

z

)

≤

f

⁢

(

x

,

1

+

max

⁡

{

y

,

z

}

)

.

8.

f

⁢

(

x

,

y

)

f

⁢

(

x

,

z

)

≤

f

⁢

(

x

,

2

+

max

⁡

{

y

,

z

}

)

.

9.

f

⁢

(

f

⁢

(

x

,

y

)

,

z

)

≤

f

⁢

(

x

,

y

+

2

⁢

z

)

.

10.

y

<

f

⁢

(

x

,

y

)

.

Proof.

Most of the proofs are done by

induction

.

1.

The case when

x

=

0

is

obvious

. Assume now that

x

≠

0

. Induct on

y

. The case

y

=

0

is clear. Suppose

x

≤

f

⁢

(

x

,

y

)

. Then

x

≤

x

x

≤

x

f

⁢

(

x

,

y

)

=

f

⁢

(

x

,

y

+

1

)

.

2.

To see

f

⁢

(

x

,

y

)

<

f

⁢

(

x

,

y

+

1

)

for

x

>

1

, induct on

y

. First,

f

⁢

(

x

,

0

)

=

x

<

x

x

=

f

⁢

(

x

,

1

)

. Next, assume

f

⁢

(

x

,

y

)

<

f

⁢

(

x

,

y

+

1

)

. Then

f

⁢

(

x

,

y

+

1

)

=

x

f

⁢

(

x

,

y

)

<

x

f

⁢

(

x

,

y

+

1

)

=

f

⁢

(

x

,

y

+

1

)

.

To see

f

⁢

(

x

,

y

)

<

f

⁢

(

x

+

1

,

y

)

for

x

>

1

, again induct on

y

. First,

f

⁢

(

x

,

0

)

=

x

<

x

+

1

=

f

⁢

(

x

+

1

,

y

)

. Next, assume

f

⁢

(

x

,

y

)

<

f

⁢

(

x

+

1

,

y

)

. Then

f

⁢

(

x

,

y

+

1

)

=

x

f

⁢

(

x

,

y

)

<

(

x

+

1

)

f

⁢

(

x

,

y

)

<

(

x

+

1

)

f

⁢

(

x

+

1

,

y

)

=

f

⁢

(

x

+

1

,

y

+

1

)

.

3.

Induct on

y

: if

y

=

0

, then

2

⁢

f

⁢

(

x

,

0

)

=

2

⁢

x

≤

x

2

≤

x

x

=

f

⁢

(

x

,

1

)

. Next, assume

2

⁢

f

⁢

(

x

,

y

)

≤

f

⁢

(

x

,

y

+

1

)

. Then

2

⁢

f

⁢

(

x

,

y

+

1

)

=

x

f

⁢

(

x

,

y

)

≤

x

2

⁢

f

⁢

(

x

,

y

)

≤

x

f

⁢

(

x

,

y

+

1

)

=

f

⁢

(

x

,

y

+

2

)

.

4.

If

y

=

0

, then

f

⁢

(

x

,

0

)

2

=

x

2

≤

x

x

=

x

f

⁢

(

x

,

0

)

=

f

⁢

(

x

,

1

)

. Otherwise,

y

=

z

+

1

. Then

f

⁢

(

x

,

y

)

2

=

f

⁢

(

x

,

z

+

1

)

2

=

x

2

⁢

f

⁢

(

x

,

z

)

≤

x

f

⁢

(

x

,

z

+

1

)

=

x

f

⁢

(

x

,

y

)

=

f

⁢

(

x

,

y

+

1

)

. The inequality

2

⁢

f

⁢

(

x

,

z

)

≤

f

⁢

(

x

,

z

+

1

)

is derived previously.

5.

If

y

=

0

, then

f

⁢

(

x

,

0

)

f

⁢

(

x

,

0

)

=

x

x

=

f

⁢

(

x

,

1

)

≤

f

⁢

(

x

,

2

)

. Otherwise,

y

=

z

+

1

. Then

f

⁢

(

x

,

y

)

f

⁢

(

x

,

y

)

=

f

⁢

(

x

,

z

+

1

)

f

⁢

(

x

,

z

+

1

)

=

x

f

⁢

(

x

,

z

)

⁢

f

⁢

(

x

,

z

+

1

)

≤

x

f

⁢

(

x

,

z

+

1

)

2

=

x

f

⁢

(

x

,

z

+

2

)

=

f

⁢

(

x

,

z

+

3

)

=

f

⁢

(

x

,

y

+

2

)

.

From the last three statements, the next three proofs can be easily settled, fisrt, let

t

=

max

⁡

{

y

,

z

}

. Then

6.

f

⁢

(

x

,

y

)

+

f

⁢

(

x

,

z

)

≤

2

⁢

f

⁢

(

x

,

t

)

≤

f

⁢

(

x

,

t

+

1

)

.

7.

f

⁢

(

x

,

y

)

⁢

f

⁢

(

x

,

z

)

=

f

⁢

(

x

,

t

)

2

≤

f

⁢

(

x

,

t

+

1

)

.

8.

f

⁢

(

x

,

y

)

f

⁢

(

x

,

z

)

≤

f

⁢

(

x

,

t

)

f

⁢

(

x

,

t

)

≤

f

⁢

(

x

,

t

+

2

)

.

9.

Induct on

z

. If

z

=

0

, then

f

⁢

(

f

⁢

(

x

,

y

)

,

0

)

=

f

⁢

(

x

,

y

)

. Next, assume

f

⁢

(

f

⁢

(

x

,

y

)

,

z

)

≤

f

⁢

(

x

,

y

+

2

⁢

z

)

. Then

f

⁢

(

f

⁢

(

x

,

y

)

,

z

+

1

)

=

f

⁢

(

x

,

y

)

f

⁢

(

f

⁢

(

x

,

y

)

,

z

)

=

f

⁢

(

x

,

y

)

f

⁢

(

x

,

y

+

2

⁢

z

)

≤

f

⁢

(

x

,

y

+

1

)

f

⁢

(

x

,

y

+

2

⁢

z

)

≤

x

f

⁢

(

x

,

y

)

⁢

f

⁢

(

x

,

y

+

2

⁢

z

)

≤

x

f

⁢

(

x

,

y

+

2

⁢

z

+

1

)

=

f

⁢

(

x

,

y

+

2

⁢

z

+

2

)

.

10.

Induct on

y

. The case when

y

=

0

is obvious. Next, if

y

<

f

⁢

(

x

,

y

)

, then

y

+

1

<

f

⁢

(

x

,

y

)

+

1

<

f

⁢

(

x

,

y

)

+

f

⁢

(

x

,

0

)

≤

f

⁢

(

x

,

y

+

1

)

.

∎

Concerning the recursiveness of

f

, here is another basic property of

f

:

Proposition 2

.

f

is

primitive recursive

.

Proof.

Since

f

⁢

(

m

,

0

)

=

m

=

p

1

1

⁢

(

m

)

and

f

⁢

(

m

,

n

+

1

)

=

exp

⁡

(

m

,

f

⁢

(

m

,

n

)

)

=

g

⁢

(

m

,

n

,

f

⁢

(

m

,

n

)

)

, where

g

⁢

(

x

,

y

,

z

)

=

exp

⁡

(

p

1

3

⁢

(

x

,

y

,

z

)

,

p

3

3

⁢

(

x

,

y

,

z

)

)

, are defined by

primitive recursion

via functions

p

1

1

and

g

, and since the projection functions

p

i

j

, the

exponential function

exp

, and consequently

g

, are primitive recursive (

g

obtained by composition), we see that

f

is primitive recursive.
∎

Remark

. Another

recursive

property of

f

is that

f

is not

elementary recursive

. The proof uses the properties listed above. It is a bit lengthy, and is done in the link below.

Title

properties of superexponentiation

Canonical

name

PropertiesOfSuperexponentiation

Date of creation

2013-03-22 19:07:21

Last modified on

2013-03-22 19:07:21

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Result

Classification

msc 40-00

Classification

msc 03D20

Related topic

SuperexponentiationIsNotElementary

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