Term Rewriting System R:
[X, Y]
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MIN(s(X), s(Y)) -> MIN(X, Y)
QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MIN(X, Y)
LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))
LOG(s(s(X))) -> QUOT(X, s(s(0)))

Furthermore, R contains three SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

MIN(s(X), s(Y)) -> MIN(X, Y)

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

The following dependency pair can be strictly oriented:

MIN(s(X), s(Y)) -> MIN(X, Y)

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(MIN(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

The following dependency pair can be strictly oriented:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

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

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(QUOT(x1, x2)) =  x1 POL(0) =  1 POL(min(x1, x2)) =  x1 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`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Narrowing Transformation`

Dependency Pair:

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

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

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))
two new Dependency Pairs are created:

LOG(s(s(0))) -> LOG(s(0))
LOG(s(s(s(X'')))) -> LOG(s(s(quot(min(X'', s(0)), s(s(0))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Narrowing Transformation`

Dependency Pair:

LOG(s(s(s(X'')))) -> LOG(s(s(quot(min(X'', s(0)), s(s(0))))))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

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

LOG(s(s(s(X'')))) -> LOG(s(s(quot(min(X'', s(0)), s(s(0))))))
one new Dependency Pair is created:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(min(X', 0), s(s(0))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Polynomial Ordering`

Dependency Pair:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(min(X', 0), s(s(0))))))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

The following dependency pair can be strictly oriented:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(min(X', 0), s(s(0))))))

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

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(min(x1, x2)) =  x1 POL(quot(x1, x2)) =  x1 POL(s(x1)) =  1 + x1 POL(LOG(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Using the Dependency Graph resulted in no new DP problems.

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