Term Rewriting System R:
[X, XS, N]
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
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:

AFTER(s(N), cons(X, XS)) -> AFTER(N, activate(XS))
AFTER(s(N), cons(X, XS)) -> ACTIVATE(XS)
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`
`         ↳Polo`

Dependency Pairs:

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

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
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`
`         ↳Polo`

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
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`
`         ↳Polo`

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
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`
`         ↳Polynomial Ordering`

Dependency Pair:

AFTER(s(N), cons(X, XS)) -> AFTER(N, activate(XS))

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
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:

AFTER(s(N), cons(X, XS)) -> AFTER(N, activate(XS))

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(AFTER(x1, x2)) =  x1 POL(cons(x1, x2)) =  0 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`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
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