Term Rewriting System R:
[x, y]
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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:

MINUS(x, s(y)) -> PRED(minus(x, y))
MINUS(x, s(y)) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(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`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

MINUS(x, s(y)) -> MINUS(x, y)

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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:

MINUS(x, s(y)) -> MINUS(x, y)

The following rules can be oriented:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(pred(x1)) =  x1 POL(log(x1)) =  x1 POL(MINUS(x1, x2)) =  1 + x1 + x2 POL(s(x1)) =  1 + x1

resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)
pred(x1) -> pred(x1)
minus(x1, x2) -> x1
quot(x1, x2) -> x1
log(x1) -> log(x1)

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

Dependency Pair:

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(minus(x, y), s(y))

The following rules can be oriented:

minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(QUOT(x1, x2)) =  1 + x1 + x2 POL(0) =  0 POL(pred(x1)) =  x1 POL(log(x1)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2) -> QUOT(x1, x2)
s(x1) -> s(x1)
minus(x1, x2) -> x1
pred(x1) -> pred(x1)
quot(x1, x2) -> x1
log(x1) -> log(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳Narrowing Transformation`

Dependency Pair:

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

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(minus(x'', s(0)), s(s(0))))))

The transformation is resulting in one new DP problem:

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

Dependency Pair:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(minus(x'', s(0)), s(s(0))))))
one new Dependency Pair is created:

LOG(s(s(s(x''')))) -> LOG(s(s(quot(pred(minus(x''', 0)), s(s(0))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Argument Filtering and Ordering`

Dependency Pair:

LOG(s(s(s(x''')))) -> LOG(s(s(quot(pred(minus(x''', 0)), s(s(0))))))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(x''')))) -> LOG(s(s(quot(pred(minus(x''', 0)), s(s(0))))))

The following rules can be oriented:

quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

resulting in one new DP problem.
Used Argument Filtering System:
LOG(x1) -> LOG(x1)
s(x1) -> s(x1)
quot(x1, x2) -> x1
pred(x1) -> pred(x1)
minus(x1, x2) -> x1
log(x1) -> log(x1)

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

Dependency Pair:

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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