Termination Proof using AProVE

Term Rewriting System R:
[X, XS, N, X1, X2]
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

2ND(cons(X, XS)) -> HEAD(activate(XS))
2ND(cons(X, XS)) -> ACTIVATE(XS)
TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)

one new Dependency Pair is created:

TAKE(s(N), cons(X, n_{_take}(X1'', X2''))) -> ACTIVATE(n_{_take}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

TAKE(s(N), cons(X, n_{_take}(X1'', X2''))) -> ACTIVATE(n_{_take}(X1'', X2''))
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)

one new Dependency Pair is created:

ACTIVATE(n_{_take}(s(N''), cons(X'', n_{_take}(X1'''', X2'''')))) -> TAKE(s(N''), cons(X'', n_{_take}(X1'''', X2'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

ACTIVATE(n_{_take}(s(N''), cons(X'', n_{_take}(X1'''', X2'''')))) -> TAKE(s(N''), cons(X'', n_{_take}(X1'''', X2'''')))
TAKE(s(N), cons(X, n_{_take}(X1'', X2''))) -> ACTIVATE(n_{_take}(X1'', X2''))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

TAKE(s(N), cons(X, n_{_take}(X1'', X2''))) -> ACTIVATE(n_{_take}(X1'', X2''))

one new Dependency Pair is created:

TAKE(s(N), cons(X, n_{_take}(s(N''''), cons(X'''', n_{_take}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_take}(s(N''''), cons(X'''', n_{_take}(X1'''''', X2''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

TAKE(s(N), cons(X, n_{_take}(s(N''''), cons(X'''', n_{_take}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_take}(s(N''''), cons(X'''', n_{_take}(X1'''''', X2''''''))))
ACTIVATE(n_{_take}(s(N''), cons(X'', n_{_take}(X1'''', X2'''')))) -> TAKE(s(N''), cons(X'', n_{_take}(X1'''', X2'''')))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

ACTIVATE(n_{_take}(s(N''), cons(X'', n_{_take}(X1'''', X2'''')))) -> TAKE(s(N''), cons(X'', n_{_take}(X1'''', X2'''')))

one new Dependency Pair is created:

ACTIVATE(n_{_take}(s(N'''), cons(X''', n_{_take}(s(N''''''), cons(X'''''', n_{_take}(X1'''''''', X2'''''''')))))) -> TAKE(s(N'''), cons(X''', n_{_take}(s(N''''''), cons(X'''''', n_{_take}(X1'''''''', X2'''''''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 6

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pairs:

ACTIVATE(n_{_take}(s(N'''), cons(X''', n_{_take}(s(N''''''), cons(X'''''', n_{_take}(X1'''''''', X2'''''''')))))) -> TAKE(s(N'''), cons(X''', n_{_take}(s(N''''''), cons(X'''''', n_{_take}(X1'''''''', X2'''''''')))))
TAKE(s(N), cons(X, n_{_take}(s(N''''), cons(X'''', n_{_take}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_take}(s(N''''), cons(X'''', n_{_take}(X1'''''', X2''''''))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_take}(s(N'''), cons(X''', n_{_take}(s(N''''''), cons(X'''''', n_{_take}(X1'''''''', X2'''''''')))))) -> TAKE(s(N'''), cons(X''', n_{_take}(s(N''''''), cons(X'''''', n_{_take}(X1'''''''', X2'''''''')))))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__take(x₁, x₂)) = 1 + x₂

POL(cons(x₁, x₂)) = x₂

POL(TAKE(x₁, x₂)) = x₂

POL(s(x₁)) = 0

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

TAKE(s(N), cons(X, n_{_take}(s(N''''), cons(X'''', n_{_take}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_take}(s(N''''), cons(X'''', n_{_take}(X1'''''', X2''''''))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pair:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))

three new Dependency Pairs are created:

SEL(s(N), cons(X, n_{_from}(X''))) -> SEL(N, from(X''))
SEL(s(N), cons(X, n_{_take}(X1', X2'))) -> SEL(N, take(X1', X2'))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Narrowing Transformation

Dependency Pairs:

SEL(s(N), cons(X, XS')) -> SEL(N, XS')
SEL(s(N), cons(X, n_{_take}(X1', X2'))) -> SEL(N, take(X1', X2'))
SEL(s(N), cons(X, n_{_from}(X''))) -> SEL(N, from(X''))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SEL(s(N), cons(X, n_{_from}(X''))) -> SEL(N, from(X''))

two new Dependency Pairs are created:

SEL(s(N), cons(X, n_{_from}(X'''))) -> SEL(N, cons(X''', n_{_from}(s(X'''))))
SEL(s(N), cons(X, n_{_from}(X'''))) -> SEL(N, n_{_from}(X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 9

                 ↳Narrowing Transformation

Dependency Pairs:

SEL(s(N), cons(X, n_{_from}(X'''))) -> SEL(N, cons(X''', n_{_from}(s(X'''))))
SEL(s(N), cons(X, n_{_take}(X1', X2'))) -> SEL(N, take(X1', X2'))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SEL(s(N), cons(X, n_{_take}(X1', X2'))) -> SEL(N, take(X1', X2'))

three new Dependency Pairs are created:

SEL(s(N), cons(X, n_{_take}(0, X2''))) -> SEL(N, nil)
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', XS')))) -> SEL(N, cons(X'', n_{_take}(N'', activate(XS'))))
SEL(s(N), cons(X, n_{_take}(X1'', X2''))) -> SEL(N, n_{_take}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 10

                 ↳Forward Instantiation Transformation

Dependency Pairs:

SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', XS')))) -> SEL(N, cons(X'', n_{_take}(N'', activate(XS'))))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')
SEL(s(N), cons(X, n_{_from}(X'''))) -> SEL(N, cons(X''', n_{_from}(s(X'''))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

SEL(s(N), cons(X, XS')) -> SEL(N, XS')

three new Dependency Pairs are created:

SEL(s(s(N'')), cons(X, cons(X'', XS'''))) -> SEL(s(N''), cons(X'', XS'''))
SEL(s(s(N'')), cons(X, cons(X'', n_{_from}(X''''')))) -> SEL(s(N''), cons(X'', n_{_from}(X''''')))
SEL(s(s(N'')), cons(X, cons(X'', n_{_take}(s(N''''), cons(X'''', XS'''))))) -> SEL(s(N''), cons(X'', n_{_take}(s(N''''), cons(X'''', XS'''))))

The transformation is resulting in three new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 11

                 ↳Polynomial Ordering

Dependency Pair:

SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', XS')))) -> SEL(N, cons(X'', n_{_take}(N'', activate(XS'))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', XS')))) -> SEL(N, cons(X'', n_{_take}(N'', activate(XS'))))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from(x₁)) = 0

POL(from(x₁)) = 0

POL(activate(x₁)) = 0

POL(0) = 0

POL(n__take(x₁, x₂)) = 0

POL(cons(x₁, x₂)) = 0

POL(SEL(x₁, x₂)) = x₁

POL(take(x₁, x₂)) = 0

POL(nil) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 14

                 ↳Dependency Graph

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 12

                 ↳Polynomial Ordering

Dependency Pair:

SEL(s(N), cons(X, n_{_from}(X'''))) -> SEL(N, cons(X''', n_{_from}(s(X'''))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, n_{_from}(X'''))) -> SEL(N, cons(X''', n_{_from}(s(X'''))))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from(x₁)) = 0

POL(SEL(x₁, x₂)) = x₁

POL(cons(x₁, x₂)) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 13

                 ↳Polynomial Ordering

Dependency Pair:

SEL(s(s(N'')), cons(X, cons(X'', XS'''))) -> SEL(s(N''), cons(X'', XS'''))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

SEL(s(s(N'')), cons(X, cons(X'', XS'''))) -> SEL(s(N''), cons(X'', XS'''))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(SEL(x₁, x₂)) = x₁

POL(cons(x₁, x₂)) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

POL(n__take(x₁, x₂))	= 1 + x₂
POL(cons(x₁, x₂))	= x₂
POL(TAKE(x₁, x₂))	= x₂
POL(s(x₁))	= 0
POL(ACTIVATE(x₁))	= x₁

POL(n__from(x₁))	= 0
POL(from(x₁))	= 0
POL(activate(x₁))	= 0
POL(0)	= 0
POL(n__take(x₁, x₂))	= 0
POL(cons(x₁, x₂))	= 0
POL(SEL(x₁, x₂))	= x₁
POL(take(x₁, x₂))	= 0
POL(nil)	= 0
POL(s(x₁))	= 1 + x₁