Term Rewriting System R:
[X, Y, Z]
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(X) -> X

Termination of R to be shown.

`   R`
`     ↳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(nfrom(X)) -> FROM(activate(X))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pairs:

ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(ns(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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 3`
`             ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

ACTIVATE(nfrom(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(nfrom(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.

`   R`
`     ↳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, nfrom(ns(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳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, nfrom(ns(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(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, nfrom(X''))) -> SEL(X, from(activate(X'')))
SEL(s(X), cons(Y, ns(X''))) -> SEL(X, s(activate(X'')))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

The transformation is resulting in one new DP problem:

`   R`
`     ↳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, ns(X''))) -> SEL(X, s(activate(X'')))
SEL(s(X), cons(Y, nfrom(X''))) -> SEL(X, from(activate(X'')))

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(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, nfrom(X''))) -> SEL(X, from(activate(X'')))
five new Dependency Pairs are created:

SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(activate(X'''), nfrom(ns(activate(X''')))))
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, nfrom(activate(X''')))
SEL(s(X), cons(Y, nfrom(nfrom(X''')))) -> SEL(X, from(from(activate(X'''))))
SEL(s(X), cons(Y, nfrom(ns(X''')))) -> SEL(X, from(s(activate(X'''))))
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, from(X'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳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, nfrom(X'''))) -> SEL(X, from(X'''))
SEL(s(X), cons(Y, nfrom(nfrom(X''')))) -> SEL(X, from(from(activate(X'''))))
SEL(s(X), cons(Y, nfrom(ns(X''')))) -> SEL(X, from(s(activate(X'''))))
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(activate(X'''), nfrom(ns(activate(X''')))))
SEL(s(X), cons(Y, ns(X''))) -> SEL(X, s(activate(X'')))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(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, ns(X''))) -> SEL(X, s(activate(X'')))
four new Dependency Pairs are created:

SEL(s(X), cons(Y, ns(X'''))) -> SEL(X, ns(activate(X''')))
SEL(s(X), cons(Y, ns(nfrom(X''')))) -> SEL(X, s(from(activate(X'''))))
SEL(s(X), cons(Y, ns(ns(X''')))) -> SEL(X, s(s(activate(X'''))))
SEL(s(X), cons(Y, ns(X'''))) -> SEL(X, s(X'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳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, ns(X'''))) -> SEL(X, s(X'''))
SEL(s(X), cons(Y, ns(ns(X''')))) -> SEL(X, s(s(activate(X'''))))
SEL(s(X), cons(Y, ns(nfrom(X''')))) -> SEL(X, s(from(activate(X'''))))
SEL(s(X), cons(Y, nfrom(nfrom(X''')))) -> SEL(X, from(from(activate(X'''))))
SEL(s(X), cons(Y, nfrom(ns(X''')))) -> SEL(X, from(s(activate(X'''))))
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(activate(X'''), nfrom(ns(activate(X''')))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, from(X'''))

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(X) -> X

The following dependency pairs can be strictly oriented:

SEL(s(X), cons(Y, ns(X'''))) -> SEL(X, s(X'''))
SEL(s(X), cons(Y, ns(ns(X''')))) -> SEL(X, s(s(activate(X'''))))
SEL(s(X), cons(Y, ns(nfrom(X''')))) -> SEL(X, s(from(activate(X'''))))
SEL(s(X), cons(Y, nfrom(nfrom(X''')))) -> SEL(X, from(from(activate(X'''))))
SEL(s(X), cons(Y, nfrom(ns(X''')))) -> SEL(X, from(s(activate(X'''))))
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(activate(X'''), nfrom(ns(activate(X''')))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, nfrom(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.

`   R`
`     ↳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, nfrom(ns(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(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