Termination Proof using AProVE

Term Rewriting System R:
[X, Z, Y, X1, X2]
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pairs:

ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)

Additionally, the following rules can be oriented:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

POL(n__from(x₁)) = x₁

POL(activate(x₁)) = x₁

POL(first(x₁, x₂)) = 1 + x₂

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(FIRST(x₁, x₂)) = x₂

POL(nil) = 0

POL(s(x₁)) = 0

POL(sel(x₁, x₂)) = 1 + x₂

POL(ACTIVATE(x₁)) = x₁

POL(n__first(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, Z)) -> SEL(X, activate(Z))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, Z)) -> SEL(X, activate(Z))

three new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_from}(X''))) -> SEL(X, from(X''))
SEL(s(X), cons(Y, n_{_first}(X1', X2'))) -> SEL(X, first(X1', X2'))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

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(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_first}(X1', X2'))) -> SEL(X, first(X1', X2'))
SEL(s(X), cons(Y, n_{_from}(X''))) -> SEL(X, from(X''))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_from}(X''))) -> SEL(X, from(X''))

two new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, 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(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(X1', X2'))) -> SEL(X, first(X1', X2'))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_first}(X1', X2'))) -> SEL(X, first(X1', X2'))

three new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(0, X2''))) -> SEL(X, nil)
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))
SEL(s(X), cons(Y, n_{_first}(X1'', X2''))) -> SEL(X, n_{_first}(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(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

Additionally, the following rules can be oriented:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(n__from(x₁)) = x₁

POL(activate(x₁)) = 1 + x₁

POL(first(x₁, x₂)) = 1 + x₂

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(SEL(x₁, x₂)) = 1 + x₂

POL(nil) = 0

POL(s(x₁)) = 0

POL(sel(x₁, x₂)) = 1 + x₂

POL(n__first(x₁, x₂)) = x₂

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(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))

three new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', from(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', first(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))

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(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', first(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', from(X'''))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', from(X'''))))

two new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', 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(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', first(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', first(X1', X2'))))

three new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_first}(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(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))

Additionally, the following rules can be oriented:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(n__from(x₁)) = x₁

POL(activate(x₁)) = 1 + x₁

POL(first(x₁, x₂)) = 1 + x₂

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(SEL(x₁, x₂)) = 1 + x₂

POL(nil) = 0

POL(s(x₁)) = 0

POL(sel(x₁, x₂)) = 1 + x₂

POL(n__first(x₁, x₂)) = x₂

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(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))

three new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', Z'')))))

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(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', Z'')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', Z'')))))

Additionally, the following rules can be oriented:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(n__from(x₁)) = x₁

POL(activate(x₁)) = 1 + x₁

POL(first(x₁, x₂)) = 1 + x₂

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(SEL(x₁, x₂)) = 1 + x₂

POL(nil) = 0

POL(s(x₁)) = 0

POL(sel(x₁, x₂)) = 1 + x₂

POL(n__first(x₁, x₂)) = x₂

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(X''''''), cons(s(s(X'''''''''''')), n_{_from}(s(s(s(X''''''''''''')))))) -> SEL(X'''''', 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)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X
Dependency Pairs:
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X
Dependency Pair:
SEL(s(X0), cons(Y', n_{_first}(s(X''0), cons(Y'''', n_{_from}(s(X''''''')))))) -> SEL(X0, cons(Y'''', n_{_first}(X''0, cons(s(X'''''''), n_{_from}(s(s(X''''''')))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

one new Dependency Pair is created:

SEL(s(X0), cons(Y', n_{_first}(s(X''0), cons(Y'''', n_{_from}(s(X''''''')))))) -> SEL(X0, cons(Y'''', n_{_first}(X''0, 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(X''''''), cons(s(s(X'''''''''''')), n_{_from}(s(s(s(X''''''''''''')))))) -> SEL(X'''''', 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)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X
Dependency Pairs:
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X
Dependency Pair:
SEL(s(X0), cons(Y', n_{_first}(s(X''0), cons(Y'''', n_{_from}(s(X''''''')))))) -> SEL(X0, cons(Y'''', n_{_first}(X''0, cons(s(X'''''''), n_{_from}(s(s(X''''''')))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))

one new Dependency Pair is created:

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

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X''), cons(Y', n_{_from}(s(X'''''')))) -> SEL(X'', cons(s(X''''''), n_{_from}(s(s(X'''''')))))

one new Dependency Pair is created:

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

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(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(X''''), cons(s(X'''''''0), n_{_from}(s(s(X'''''''''))))) -> SEL(X'''', cons(s(s(X''''''''')), n_{_from}(s(s(s(X'''''''''))))))

one new Dependency Pair is created:

SEL(s(X''''''), cons(s(s(X'''''''''''')), n_{_from}(s(s(s(X''''''''''''')))))) -> SEL(X'''''', 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(X''''''), cons(s(s(X'''''''''''')), n_{_from}(s(s(s(X''''''''''''')))))) -> SEL(X'''''', 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)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X
Dependency Pairs:
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X
Dependency Pair:
SEL(s(X0), cons(Y', n_{_first}(s(X''0), cons(Y'''', n_{_from}(s(X''''''')))))) -> SEL(X0, cons(Y'''', n_{_first}(X''0, cons(s(X'''''''), n_{_from}(s(s(X''''''')))))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(n_{_from}(X)) -> from(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

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

POL(from(x₁))	= x₁
POL(n__from(x₁))	= x₁
POL(activate(x₁))	= x₁
POL(first(x₁, x₂))	= 1 + x₂
POL(0)	= 0
POL(cons(x₁, x₂))	= x₁ + x₂
POL(FIRST(x₁, x₂))	= x₂
POL(nil)	= 0
POL(s(x₁))	= 0
POL(sel(x₁, x₂))	= 1 + x₂
POL(ACTIVATE(x₁))	= x₁
POL(n__first(x₁, x₂))	= 1 + x₂

POL(from(x₁))	= 1 + x₁
POL(n__from(x₁))	= x₁
POL(activate(x₁))	= 1 + x₁
POL(first(x₁, x₂))	= 1 + x₂
POL(0)	= 0
POL(cons(x₁, x₂))	= 1 + x₁ + x₂
POL(SEL(x₁, x₂))	= 1 + x₂
POL(nil)	= 0
POL(s(x₁))	= 0
POL(sel(x₁, x₂))	= 1 + x₂
POL(n__first(x₁, x₂))	= x₂