Term Rewriting System R:
[X, Z, Y, X1, X2]
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Termination of R to be shown.

`   R`
`     ↳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(nfrom(X)) -> FROM(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)

Furthermore, R contains two SCCs.

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

Dependency Pairs:

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(nfirst(X1, X2)) -> FIRST(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)) =  x1 + x2 POL(FIRST(x1, x2)) =  x2 POL(s(x1)) =  0 POL(ACTIVATE(x1)) =  x1 POL(n__first(x1, x2)) =  1 + x2

resulting in one new DP problem.

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

Dependency Pair:

FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(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(X), cons(Y, Z)) -> SEL(X, activate(Z))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(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, nfrom(X''))) -> SEL(X, from(X''))
SEL(s(X), cons(Y, nfirst(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:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →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, nfirst(X1', X2'))) -> SEL(X, first(X1', X2'))
SEL(s(X), cons(Y, nfrom(X''))) -> SEL(X, from(X''))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(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, nfrom(X''))) -> SEL(X, from(X''))
two new Dependency Pairs are created:

SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, 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(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))
SEL(s(X), cons(Y, nfirst(X1', X2'))) -> SEL(X, first(X1', X2'))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(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, nfirst(X1', X2'))) -> SEL(X, first(X1', X2'))
three new Dependency Pairs are created:

SEL(s(X), cons(Y, nfirst(0, X2''))) -> SEL(X, nil)
SEL(s(X), cons(Y, nfirst(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', nfirst(X'', activate(Z'))))
SEL(s(X), cons(Y, nfirst(X1'', X2''))) -> SEL(X, nfirst(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(X), cons(Y, nfirst(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', nfirst(X'', activate(Z'))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pairs can be strictly oriented:

SEL(s(X), cons(Y, nfirst(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', nfirst(X'', activate(Z'))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, 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(first(x1, x2)) =  0 POL(0) =  0 POL(cons(x1, x2)) =  0 POL(SEL(x1, x2)) =  x1 POL(nil) =  0 POL(s(x1)) =  1 + x1 POL(n__first(x1, x2)) =  0

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)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(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