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

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

         ↳Polynomial Ordering

       →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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X
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))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

POL(activate(x₁)) = x₁

POL(n__take(x₁, x₂)) = 1 + x₂

POL(take(x₁, x₂)) = 1 + x₂

POL(TAKE(x₁, x₂)) = x₂

POL(sel(x₁, x₂)) = 1 + x₂

POL(ACTIVATE(x₁)) = x₁

POL(n__from(x₁)) = x₁

POL(0) = 0

POL(2nd(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁ + x₂

POL(nil) = 0

POL(s(x₁)) = 0

POL(head(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →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

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

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳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

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

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 5

                 ↳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

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

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 6

                 ↳Polynomial Ordering

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X
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))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(activate(x₁)) = 1 + x₁

POL(n__take(x₁, x₂)) = x₂

POL(SEL(x₁, x₂)) = 1 + x₂

POL(take(x₁, x₂)) = 1 + x₂

POL(sel(x₁, x₂)) = 1 + x₂

POL(n__from(x₁)) = x₁

POL(0) = 0

POL(2nd(x₁)) = x₁

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(nil) = 0

POL(s(x₁)) = 0

POL(head(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 7

                 ↳Dependency Graph

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, 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

Using the Dependency Graph the DP problem was split into 2 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 8

                 ↳Narrowing Transformation

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 10

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 12

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 14

                 ↳Polynomial Ordering

Dependency Pairs:

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

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'', XS'')))

Additionally, the following rules can be oriented:

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))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X
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))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(activate(x₁)) = 1 + x₁

POL(n__take(x₁, x₂)) = x₂

POL(SEL(x₁, x₂)) = 1 + x₂

POL(take(x₁, x₂)) = 1 + x₂

POL(sel(x₁, x₂)) = 1 + x₂

POL(n__from(x₁)) = x₁

POL(0) = 0

POL(2nd(x₁)) = x₁

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(nil) = 0

POL(s(x₁)) = 0

POL(head(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 16

                 ↳Dependency Graph

Dependency Pairs:

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

Using the Dependency Graph the DP problem was split into 2 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 17

                 ↳Narrowing Transformation

Dependency Pair:

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

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

three new Dependency Pairs are created:

SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', n_{_from}(X'0))))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', from(X'0))))))
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', n_{_take}(X1', X2'))))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', take(X1', X2'))))))
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', XS'')))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', XS'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 19

                 ↳Polynomial Ordering

Dependency Pairs:

SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', XS'')))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', XS'')))))
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', n_{_take}(X1', X2'))))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', take(X1', X2'))))))
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', n_{_from}(X'0))))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', from(X'0))))))

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X
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))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(activate(x₁)) = 1 + x₁

POL(n__take(x₁, x₂)) = x₂

POL(SEL(x₁, x₂)) = 1 + x₂

POL(take(x₁, x₂)) = 1 + x₂

POL(sel(x₁, x₂)) = 1 + x₂

POL(n__from(x₁)) = x₁

POL(0) = 0

POL(2nd(x₁)) = x₁

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(nil) = 0

POL(s(x₁)) = 0

POL(head(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 20

                 ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
SEL(s(N''''''), cons(s(s(X'''''''''''')), n_{_from}(s(s(s(X''''''''''''')))))) -> SEL(N'''''', cons(s(s(s(X'''''''''''''))), n_{_from}(s(s(s(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
Dependency Pairs:
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', n_{_take}(X1', X2'))))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', take(X1', X2'))))))
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', n_{_from}(X'0))))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', from(X'0))))))

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
Dependency Pair:
SEL(s(N0), cons(X', n_{_take}(s(N''''), cons(X''0, n_{_from}(s(X''''''')))))) -> SEL(N0, cons(X''0, n_{_take}(N'''', cons(s(X'''''''), n_{_from}(s(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

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 18

                 ↳Instantiation Transformation

Dependency Pair:

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

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

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

one new Dependency Pair is created:

SEL(s(N0), cons(X', n_{_take}(s(N''''), cons(X''0, n_{_from}(s(X''''''')))))) -> SEL(N0, cons(X''0, n_{_take}(N'''', cons(s(X'''''''), n_{_from}(s(s(X''''''')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 20

                 ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
SEL(s(N''''''), cons(s(s(X'''''''''''')), n_{_from}(s(s(s(X''''''''''''')))))) -> SEL(N'''''', cons(s(s(s(X'''''''''''''))), n_{_from}(s(s(s(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
Dependency Pairs:
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', n_{_take}(X1', X2'))))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', take(X1', X2'))))))
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', n_{_from}(X'0))))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', from(X'0))))))

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
Dependency Pair:
SEL(s(N0), cons(X', n_{_take}(s(N''''), cons(X''0, n_{_from}(s(X''''''')))))) -> SEL(N0, cons(X''0, n_{_take}(N'''', cons(s(X'''''''), n_{_from}(s(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

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 9

                 ↳Instantiation Transformation

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 11

                 ↳Instantiation Transformation

Dependency Pair:

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

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

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

one new Dependency Pair is created:

SEL(s(N''''), cons(s(X'''''''0), n_{_from}(s(s(X'''''''''))))) -> SEL(N'''', cons(s(s(X''''''''')), n_{_from}(s(s(s(X'''''''''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 13

                 ↳Instantiation Transformation

Dependency Pair:

SEL(s(N''''), cons(s(X'''''''0), n_{_from}(s(s(X'''''''''))))) -> SEL(N'''', cons(s(s(X''''''''')), n_{_from}(s(s(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

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

SEL(s(N''''), cons(s(X'''''''0), n_{_from}(s(s(X'''''''''))))) -> SEL(N'''', cons(s(s(X''''''''')), n_{_from}(s(s(s(X'''''''''))))))

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 20

                 ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
SEL(s(N''''''), cons(s(s(X'''''''''''')), n_{_from}(s(s(s(X''''''''''''')))))) -> SEL(N'''''', cons(s(s(s(X'''''''''''''))), n_{_from}(s(s(s(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
Dependency Pairs:
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', n_{_take}(X1', X2'))))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', take(X1', X2'))))))
SEL(s(N), cons(X, n_{_take}(s(N''), cons(X'', n_{_take}(s(N'''), cons(X''', n_{_from}(X'0))))))) -> SEL(N, cons(X'', n_{_take}(N'', cons(X''', n_{_take}(N''', from(X'0))))))

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
Dependency Pair:
SEL(s(N0), cons(X', n_{_take}(s(N''''), cons(X''0, n_{_from}(s(X''''''')))))) -> SEL(N0, cons(X''0, n_{_take}(N'''', cons(s(X'''''''), n_{_from}(s(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

Termination of R could not be shown.
Duration:
0:01 minutes

POL(from(x₁))	= x₁
POL(activate(x₁))	= x₁
POL(n__take(x₁, x₂))	= 1 + x₂
POL(take(x₁, x₂))	= 1 + x₂
POL(TAKE(x₁, x₂))	= x₂
POL(sel(x₁, x₂))	= 1 + x₂
POL(ACTIVATE(x₁))	= x₁
POL(n__from(x₁))	= x₁
POL(0)	= 0
POL(2nd(x₁))	= x₁
POL(cons(x₁, x₂))	= x₁ + x₂
POL(nil)	= 0
POL(s(x₁))	= 0
POL(head(x₁))	= x₁

POL(from(x₁))	= 1 + x₁
POL(activate(x₁))	= 1 + x₁
POL(n__take(x₁, x₂))	= x₂
POL(SEL(x₁, x₂))	= 1 + x₂
POL(take(x₁, x₂))	= 1 + x₂
POL(sel(x₁, x₂))	= 1 + x₂
POL(n__from(x₁))	= x₁
POL(0)	= 0
POL(2nd(x₁))	= x₁
POL(cons(x₁, x₂))	= 1 + x₁ + x₂
POL(nil)	= 0
POL(s(x₁))	= 0
POL(head(x₁))	= x₁