Term Rewriting System R:
[X, XS, N, X1, X2]
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(X) -> X

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

2ND(cons(X, XS)) -> ACTIVATE(XS)
TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(nfrom(X)) -> FROM(X)
ACTIVATE(ntake(X1, X2)) -> TAKE(X1, X2)

Furthermore, R contains two SCCs.

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

Dependency Pairs:

TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(ntake(X1, X2)) -> TAKE(X1, X2)

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(ntake(X1, X2)) -> TAKE(X1, X2)

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(cons(x1, x2)) =  x2 POL(n__take(x1, x2)) =  1 + x2 POL(TAKE(x1, x2)) =  x2 POL(s(x1)) =  x1 POL(ACTIVATE(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 3`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(X1, X2)
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(N), cons(X, XS)) -> SEL(N, activate(XS))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(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(N), cons(X, XS)) -> SEL(N, activate(XS))
three new Dependency Pairs are created:

SEL(s(N), cons(X, nfrom(X''))) -> SEL(N, from(X''))
SEL(s(N), cons(X, ntake(X1', X2'))) -> SEL(N, take(X1', X2'))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Narrowing Transformation`

Dependency Pairs:

SEL(s(N), cons(X, XS')) -> SEL(N, XS')
SEL(s(N), cons(X, ntake(X1', X2'))) -> SEL(N, take(X1', X2'))
SEL(s(N), cons(X, nfrom(X''))) -> SEL(N, from(X''))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(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(N), cons(X, nfrom(X''))) -> SEL(N, from(X''))
two new Dependency Pairs are created:

SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, cons(X''', nfrom(s(X'''))))
SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, nfrom(X'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 5`
`                 ↳Narrowing Transformation`

Dependency Pairs:

SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, cons(X''', nfrom(s(X'''))))
SEL(s(N), cons(X, ntake(X1', X2'))) -> SEL(N, take(X1', X2'))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(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(N), cons(X, ntake(X1', X2'))) -> SEL(N, take(X1', X2'))
three new Dependency Pairs are created:

SEL(s(N), cons(X, ntake(0, X2''))) -> SEL(N, nil)
SEL(s(N), cons(X, ntake(s(N''), cons(X'', XS')))) -> SEL(N, cons(X'', ntake(N'', activate(XS'))))
SEL(s(N), cons(X, ntake(X1'', X2''))) -> SEL(N, ntake(X1'', X2''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 6`
`                 ↳Polynomial Ordering`

Dependency Pairs:

SEL(s(N), cons(X, ntake(s(N''), cons(X'', XS')))) -> SEL(N, cons(X'', ntake(N'', activate(XS'))))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')
SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, cons(X''', nfrom(s(X'''))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(X) -> X

The following dependency pairs can be strictly oriented:

SEL(s(N), cons(X, ntake(s(N''), cons(X'', XS')))) -> SEL(N, cons(X'', ntake(N'', activate(XS'))))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')
SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, cons(X''', nfrom(s(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(0) =  0 POL(n__take(x1, x2)) =  0 POL(cons(x1, x2)) =  0 POL(SEL(x1, x2)) =  x1 POL(take(x1, x2)) =  0 POL(nil) =  0 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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