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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

HALF(s(s(x))) -> HALF(x)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 4`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Narrowing Transformation`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Polynomial Ordering`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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