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

         ↳Polynomial 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 pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from(x₁)) = x₁

POL(n__s(x₁)) = 1 + x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pair:

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 pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from(x₁)) = 1 + x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Polo

...

               →DP Problem 4

                 ↳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

         ↳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}(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

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳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

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Nar

...

               →DP Problem 6

                 ↳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

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Nar

...

               →DP Problem 7

                 ↳Polynomial 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'''))

There are no usable rules 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(cons(x₁, x₂)) = 0

POL(SEL(x₁, x₂)) = x₁

POL(n__s(x₁)) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Nar

...

               →DP Problem 8

                 ↳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:
0:00 minutes

POL(n__from(x₁))	= x₁
POL(n__s(x₁))	= 1 + x₁
POL(ACTIVATE(x₁))	= x₁

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