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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(s(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)
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(x, s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x, 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(s(x), s(y)) -> MINUS(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)

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

Dependency Pair:

Rules:

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

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

The following dependency pair can be strictly oriented:

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

The following usable rules w.r.t. to the AFS can be oriented:

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

Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2) -> QUOT(x1, x2)
s(x1) -> s(x1)
minus(x1, x2) -> x1

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

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, 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 Pairs:

PLUS(minus(x, s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(x, s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x, s(0)))
two new Dependency Pairs are created:

PLUS(minus(x, s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x, s(0)))
PLUS(minus(s(x''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x'', 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 Pairs:

PLUS(minus(s(x''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(x, s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x, s(0)))
two new Dependency Pairs are created:

PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 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`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x'', 0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(s(x''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x'', 0))
two new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')

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 8`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')
PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
three new Dependency Pairs are created:

PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 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 9`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))
two new Dependency Pairs are created:

PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')

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 10`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')
PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
two new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')

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 11`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

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

PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')

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 12`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
three new Dependency Pairs are created:

PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 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 13`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
three new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')

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 14`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
three new Dependency Pairs are created:

PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')

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 15`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
three new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')

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 16`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
three new Dependency Pairs are created:

PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 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 17`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
three new Dependency Pairs are created:

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')

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 18`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
three new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')

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 19`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
three new Dependency Pairs are created:

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')

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 20`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

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

PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
three new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')

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 21`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))

Termination of R could not be shown.
Duration:
0:06 minutes