Term Rewriting System R:
[x]
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

HALF(s(s(x))) -> S(half(x))
HALF(s(s(x))) -> HALF(x)
S(log(0)) -> S(0)
LOG(s(x)) -> S(log(half(s(x))))
LOG(s(x)) -> LOG(half(s(x)))
LOG(s(x)) -> HALF(s(x))

Furthermore, R contains two SCCs.

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

Dependency Pair:

HALF(s(s(x))) -> HALF(x)

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))

The following dependency pair can be strictly oriented:

HALF(s(s(x))) -> HALF(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(HALF(x1)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

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

Dependency Pair:

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

LOG(s(x)) -> LOG(half(s(x)))

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

LOG(s(x)) -> LOG(half(s(x)))
three new Dependency Pairs are created:

LOG(s(0)) -> LOG(0)
LOG(s(s(x''))) -> LOG(s(half(x'')))
LOG(s(log(0))) -> LOG(half(s(0)))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

LOG(s(log(0))) -> LOG(half(s(0)))
LOG(s(s(x''))) -> LOG(s(half(x'')))

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))

The following dependency pair can be strictly oriented:

LOG(s(log(0))) -> LOG(half(s(0)))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

s(log(0)) -> s(0)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(log(x1)) =  1 POL(s(x1)) =  x1 POL(half(x1)) =  0 POL(LOG(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Polo`
`             ...`
`               →DP Problem 5`
`                 ↳Polynomial Ordering`

Dependency Pair:

LOG(s(s(x''))) -> LOG(s(half(x'')))

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))

The following dependency pair can be strictly oriented:

LOG(s(s(x''))) -> LOG(s(half(x'')))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

s(log(0)) -> s(0)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(log(x1)) =  0 POL(s(x1)) =  1 + x1 POL(half(x1)) =  x1 POL(LOG(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Polo`
`             ...`
`               →DP Problem 6`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))

Using the Dependency Graph resulted in no new DP problems.

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