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

         ↳Argument Filtering and 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 pairs can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

ACTIVATE(x₁) -> x₁
n_{_first}(x₁, x₂) -> n_{_first}(x₁, x₂)
FIRST(x₁, x₂) -> x₂
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

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

         ↳AFS

       →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

         ↳AFS

       →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

         ↳AFS

       →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

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 6

                 ↳Argument Filtering and 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 pairs can be strictly oriented:

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'''))))

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

SEL(x₁, x₂) -> SEL(x₁, x₂)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> x₂
n_{_first}(x₁, x₂) -> x₁
n_{_from}(x₁) -> n_{_from}

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 7

                 ↳Dependency Graph

Dependency Pair:

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.

Termination of R successfully shown.
Duration:
0:00 minutes