Term Rewriting System R:
[X]
f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(s(0)) -> F(p(s(0)))
F(s(0)) -> P(s(0))
ACTIVATE(nf(X)) -> F(activate(X))
ACTIVATE(nf(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(n0) -> 0'

Furthermore, R contains two SCCs.

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

Dependency Pair:

F(s(0)) -> F(p(s(0)))

Rules:

f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X

The following dependency pair can be strictly oriented:

F(s(0)) -> F(p(s(0)))

Additionally, the following usable rules using the Ce-refinement can be oriented:

p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(n__s(x1)) =  0 POL(s(x1)) =  1 POL(n__0) =  0 POL(F(x1)) =  x1 POL(p(x1)) =  0

resulting in one new DP problem.

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

Dependency Pair:

Rules:

f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
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 Pairs:

ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nf(X)) -> ACTIVATE(X)

Rules:

f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X

The following dependency pair can be strictly oriented:

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

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(n__f(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 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Polynomial Ordering`

Dependency Pair:

ACTIVATE(nf(X)) -> ACTIVATE(X)

Rules:

f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(nf(X)) -> ACTIVATE(X)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(n__f(x1)) =  1 + x1 POL(ACTIVATE(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Polo`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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