Termination Proof using AProVE

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

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_from}(X)) -> FROM(activate(X))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_s}(X)) -> S(activate(X))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

Dependency Pairs:

ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(X) -> X

The following dependency pairs can be strictly oriented:

ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

ACTIVATE(x₁) -> ACTIVATE(x₁)
n_{_s}(x₁) -> n_{_s}(x₁)
n_{_from}(x₁) -> n_{_from}(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
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}(n_{_s}(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
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(activate(X'')))
SEL(s(X), cons(Y, n_{_s}(X''))) -> SEL(X, s(activate(X'')))
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_{_s}(X''))) -> SEL(X, s(activate(X'')))
SEL(s(X), cons(Y, n_{_from}(X''))) -> SEL(X, from(activate(X'')))

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
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(activate(X'')))

five new Dependency Pairs are created:

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

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
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_{_s}(X''))) -> SEL(X, s(activate(X'')))

four new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_s}(X'''))) -> SEL(X, n_{_s}(activate(X''')))
SEL(s(X), cons(Y, n_{_s}(n_{_from}(X''')))) -> SEL(X, s(from(activate(X'''))))
SEL(s(X), cons(Y, n_{_s}(n_{_s}(X''')))) -> SEL(X, s(s(activate(X'''))))
SEL(s(X), cons(Y, n_{_s}(X'''))) -> SEL(X, s(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 6

                 ↳Argument Filtering and Ordering

Dependency Pairs:

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

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(X) -> X

The following dependency pairs can be strictly oriented:

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

The following usable rules w.r.t. to the AFS can be oriented:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(X) -> X
s(X) -> n_{_s}(X)

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

SEL > activate > from > n_{_s}
SEL > activate > from > n_{_from}
SEL > activate > from > cons
SEL > activate > s > n_{_s}

resulting in one new DP problem.
Used Argument Filtering System:

SEL(x₁, x₂) -> SEL(x₁, x₂)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
n_{_from}(x₁) -> n_{_from}(x₁)
from(x₁) -> from(x₁)
activate(x₁) -> activate(x₁)
n_{_s}(x₁) -> n_{_s}(x₁)

     ↳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}(n_{_s}(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:02 minutes