Term Rewriting System R:
[x, y, z]
minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(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(p(s(x)), p(s(y)))
MINUS(s(x), s(y)) -> P(s(x))
MINUS(s(x), s(y)) -> P(s(y))
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(x, plus(y, z)) -> MINUS(x, y)
P(s(s(x))) -> P(s(x))
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))
DIV(s(x), s(y)) -> MINUS(x, y)
DIV(plus(x, y), z) -> PLUS(div(x, z), div(y, z))
DIV(plus(x, y), z) -> DIV(x, z)
DIV(plus(x, y), z) -> DIV(y, z)
PLUS(s(x), y) -> PLUS(y, minus(s(x), s(0)))
PLUS(s(x), y) -> MINUS(s(x), s(0))

Furthermore, R contains four SCCs.

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

Dependency Pair:

P(s(s(x))) -> P(s(x))

Rules:

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

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(x, plus(y, z)) -> MINUS(x, y)
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))

Rules:

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

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

MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(x), s(s(x''))) -> MINUS(p(s(x)), s(p(s(x''))))
MINUS(s(s(x'')), s(y)) -> MINUS(s(p(s(x''))), p(s(y)))
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(x, plus(y, z)) -> MINUS(x, y)

Rules:

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

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

MINUS(s(s(x'')), s(y)) -> MINUS(s(p(s(x''))), p(s(y)))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(x'')), s(s(x'))) -> MINUS(s(p(s(x''))), s(p(s(x'))))
MINUS(x, plus(y, z)) -> MINUS(x, y)
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(s(s(s(x'))), s(y)) -> MINUS(s(s(p(s(x')))), p(s(y)))
MINUS(s(x), s(s(x''))) -> MINUS(p(s(x)), s(p(s(x''))))

Rules:

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

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

MINUS(s(x), s(s(x''))) -> MINUS(p(s(x)), s(p(s(x''))))
two new Dependency Pairs are created:

MINUS(s(s(x''')), s(s(x''))) -> MINUS(s(p(s(x'''))), s(p(s(x''))))
MINUS(s(x), s(s(s(x''')))) -> MINUS(p(s(x)), s(s(p(s(x''')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(x), s(s(s(x''')))) -> MINUS(p(s(x)), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(x''))) -> MINUS(s(p(s(x'''))), s(p(s(x''))))
MINUS(x, plus(y, z)) -> MINUS(x, y)
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(s(s(s(x'))), s(y)) -> MINUS(s(s(p(s(x')))), p(s(y)))
MINUS(s(s(x'')), s(s(x'))) -> MINUS(s(p(s(x''))), s(p(s(x'))))

Rules:

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

The following dependency pairs can be strictly oriented:

MINUS(x, plus(y, z)) -> MINUS(x, y)
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)

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

p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(plus(x1, x2)) =  1 + x1 + x2 POL(0) =  0 POL(minus(x1, x2)) =  0 POL(MINUS(x1, x2)) =  x2 POL(s(x1)) =  0 POL(p(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 9`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

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

Rules:

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

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

MINUS(s(s(s(x'))), s(y)) -> MINUS(s(s(p(s(x')))), p(s(y)))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 10`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

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

Rules:

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

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

MINUS(s(s(x'')), s(s(x'))) -> MINUS(s(p(s(x''))), s(p(s(x'))))
two new Dependency Pairs are created:

MINUS(s(s(s(x0))), s(s(x'))) -> MINUS(s(s(p(s(x0)))), s(p(s(x'))))
MINUS(s(s(x'')), s(s(s(x0)))) -> MINUS(s(p(s(x''))), s(s(p(s(x0)))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 11`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(x'')), s(s(s(x0)))) -> MINUS(s(p(s(x''))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(x'))) -> MINUS(s(s(p(s(x0)))), s(p(s(x'))))
MINUS(s(s(s(s(x'')))), s(y)) -> MINUS(s(s(s(p(s(x''))))), p(s(y)))
MINUS(s(x), s(s(s(x''')))) -> MINUS(p(s(x)), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(x''))) -> MINUS(s(p(s(x'''))), s(p(s(x''))))
MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))

Rules:

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

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

MINUS(s(s(x''')), s(s(x''))) -> MINUS(s(p(s(x'''))), s(p(s(x''))))
two new Dependency Pairs are created:

MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
MINUS(s(s(x''')), s(s(s(x')))) -> MINUS(s(p(s(x'''))), s(s(p(s(x')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 12`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(x''')), s(s(s(x')))) -> MINUS(s(p(s(x'''))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
MINUS(s(s(s(x0))), s(s(x'))) -> MINUS(s(s(p(s(x0)))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(y)) -> MINUS(s(s(s(p(s(x''))))), p(s(y)))
MINUS(s(x), s(s(s(x''')))) -> MINUS(p(s(x)), s(s(p(s(x''')))))
MINUS(s(s(x'')), s(s(s(x0)))) -> MINUS(s(p(s(x''))), s(s(p(s(x0)))))

Rules:

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

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

MINUS(s(x), s(s(s(x''')))) -> MINUS(p(s(x)), s(s(p(s(x''')))))
two new Dependency Pairs are created:

MINUS(s(s(x'')), s(s(s(x''')))) -> MINUS(s(p(s(x''))), s(s(p(s(x''')))))
MINUS(s(x), s(s(s(s(x''))))) -> MINUS(p(s(x)), s(s(s(p(s(x''))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 13`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(x), s(s(s(s(x''))))) -> MINUS(p(s(x)), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(x''')))) -> MINUS(s(p(s(x''))), s(s(p(s(x''')))))
MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(x0)))) -> MINUS(s(p(s(x''))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(x'))) -> MINUS(s(s(p(s(x0)))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(y)) -> MINUS(s(s(s(p(s(x''))))), p(s(y)))
MINUS(s(s(x''')), s(s(s(x')))) -> MINUS(s(p(s(x'''))), s(s(p(s(x')))))

Rules:

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

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

MINUS(s(s(s(s(x'')))), s(y)) -> MINUS(s(s(s(p(s(x''))))), p(s(y)))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 14`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(y)) -> MINUS(s(s(s(s(p(s(x')))))), p(s(y)))
MINUS(s(s(x'')), s(s(s(x''')))) -> MINUS(s(p(s(x''))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(x')))) -> MINUS(s(p(s(x'''))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(x0)))) -> MINUS(s(p(s(x''))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(x'))) -> MINUS(s(s(p(s(x0)))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
MINUS(s(x), s(s(s(s(x''))))) -> MINUS(p(s(x)), s(s(s(p(s(x''))))))

Rules:

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

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

MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 15`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(s(s(x'))))), s(y)) -> MINUS(s(s(s(s(p(s(x')))))), p(s(y)))
MINUS(s(x), s(s(s(s(x''))))) -> MINUS(p(s(x)), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(x''')))) -> MINUS(s(p(s(x''))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(x')))) -> MINUS(s(p(s(x'''))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(x0)))) -> MINUS(s(p(s(x''))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(x'))) -> MINUS(s(s(p(s(x0)))), s(p(s(x'))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))

Rules:

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

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

MINUS(s(s(s(x0))), s(s(x'))) -> MINUS(s(s(p(s(x0)))), s(p(s(x'))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 16`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(y)) -> MINUS(s(s(s(s(p(s(x')))))), p(s(y)))
MINUS(s(x), s(s(s(s(x''))))) -> MINUS(p(s(x)), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(x''')))) -> MINUS(s(p(s(x''))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(x')))) -> MINUS(s(p(s(x'''))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(x0)))) -> MINUS(s(p(s(x''))), s(s(p(s(x0)))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))

Rules:

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

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

MINUS(s(s(x'')), s(s(s(x0)))) -> MINUS(s(p(s(x''))), s(s(p(s(x0)))))
two new Dependency Pairs are created:

MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 17`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(y)) -> MINUS(s(s(s(s(p(s(x')))))), p(s(y)))
MINUS(s(x), s(s(s(s(x''))))) -> MINUS(p(s(x)), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(x''')))) -> MINUS(s(p(s(x''))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(x')))) -> MINUS(s(p(s(x'''))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))

Rules:

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

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

MINUS(s(s(s(x'))), s(s(x''))) -> MINUS(s(s(p(s(x')))), s(p(s(x''))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 18`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(y)) -> MINUS(s(s(s(s(p(s(x')))))), p(s(y)))
MINUS(s(x), s(s(s(s(x''))))) -> MINUS(p(s(x)), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(x''')))) -> MINUS(s(p(s(x''))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(x')))) -> MINUS(s(p(s(x'''))), s(s(p(s(x')))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))

Rules:

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

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

MINUS(s(s(x''')), s(s(s(x')))) -> MINUS(s(p(s(x'''))), s(s(p(s(x')))))
two new Dependency Pairs are created:

MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 19`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(y)) -> MINUS(s(s(s(s(p(s(x')))))), p(s(y)))
MINUS(s(x), s(s(s(s(x''))))) -> MINUS(p(s(x)), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(x''')))) -> MINUS(s(p(s(x''))), s(s(p(s(x''')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))

Rules:

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

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

MINUS(s(s(x'')), s(s(s(x''')))) -> MINUS(s(p(s(x''))), s(s(p(s(x''')))))
two new Dependency Pairs are created:

MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 20`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(y)) -> MINUS(s(s(s(s(p(s(x')))))), p(s(y)))
MINUS(s(x), s(s(s(s(x''))))) -> MINUS(p(s(x)), s(s(s(p(s(x''))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))

Rules:

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

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

MINUS(s(x), s(s(s(s(x''))))) -> MINUS(p(s(x)), s(s(s(p(s(x''))))))
two new Dependency Pairs are created:

MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 21`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(y)) -> MINUS(s(s(s(s(p(s(x')))))), p(s(y)))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))

Rules:

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

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

MINUS(s(s(s(s(s(x'))))), s(y)) -> MINUS(s(s(s(s(p(s(x')))))), p(s(y)))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 22`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))

Rules:

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

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

MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 23`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))

Rules:

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

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

MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 24`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))

Rules:

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

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

MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 25`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))

Rules:

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

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

MINUS(s(s(s(s(x'')))), s(s(x'))) -> MINUS(s(s(s(p(s(x''))))), s(p(s(x'))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 26`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))

Rules:

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

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

MINUS(s(s(s(x0))), s(s(s(x'')))) -> MINUS(s(s(p(s(x0)))), s(s(p(s(x'')))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 27`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))

Rules:

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

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

MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 28`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))

Rules:

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

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

MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
two new Dependency Pairs are created:

MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 29`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))

Rules:

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

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

MINUS(s(s(s(s(x0)))), s(s(x''))) -> MINUS(s(s(s(p(s(x0))))), s(p(s(x''))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 30`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))

Rules:

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

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

MINUS(s(s(s(x'))), s(s(s(x0)))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x0)))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 31`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))

Rules:

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

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

MINUS(s(s(s(x''))), s(s(s(x')))) -> MINUS(s(s(p(s(x'')))), s(s(p(s(x')))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(x''))), s(s(s(s(x0))))) -> MINUS(s(s(p(s(x'')))), s(s(s(p(s(x0))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 32`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(x''))), s(s(s(s(x0))))) -> MINUS(s(s(p(s(x'')))), s(s(s(p(s(x0))))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))

Rules:

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

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

MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
two new Dependency Pairs are created:

MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 33`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
MINUS(s(s(s(x''))), s(s(s(s(x0))))) -> MINUS(s(s(p(s(x'')))), s(s(s(p(s(x0))))))

Rules:

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

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

MINUS(s(s(s(x'))), s(s(s(x''')))) -> MINUS(s(s(p(s(x')))), s(s(p(s(x''')))))
two new Dependency Pairs are created:

MINUS(s(s(s(s(x'')))), s(s(s(x''')))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x''')))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 34`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x''')))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x''')))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(s(x0))))) -> MINUS(s(s(p(s(x'')))), s(s(s(p(s(x0))))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))

Rules:

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

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

MINUS(s(s(x'')), s(s(s(s(x'))))) -> MINUS(s(p(s(x''))), s(s(s(p(s(x'))))))
two new Dependency Pairs are created:

MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 35`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x'')))), s(s(s(x''')))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(s(x0))))) -> MINUS(s(s(p(s(x'')))), s(s(s(p(s(x0))))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))

Rules:

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

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

MINUS(s(s(x''')), s(s(s(s(x''))))) -> MINUS(s(p(s(x'''))), s(s(s(p(s(x''))))))
two new Dependency Pairs are created:

MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 36`
`                 ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x''')))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(s(x0))))) -> MINUS(s(s(p(s(x'')))), s(s(s(p(s(x0))))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))

Rules:

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

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

MINUS(s(x), s(s(s(s(s(x''')))))) -> MINUS(p(s(x)), s(s(s(s(p(s(x''')))))))
two new Dependency Pairs are created:

MINUS(s(s(x'')), s(s(s(s(s(x''')))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x''')))))))
MINUS(s(x), s(s(s(s(s(s(x''))))))) -> MINUS(p(s(x)), s(s(s(s(s(p(s(x''))))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pairs:

MINUS(s(x), s(s(s(s(s(s(x''))))))) -> MINUS(p(s(x)), s(s(s(s(s(p(s(x''))))))))
MINUS(s(s(x'')), s(s(s(s(s(x''')))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x''')))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x''')))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(s(x0))))) -> MINUS(s(s(p(s(x'')))), s(s(s(p(s(x0))))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))

Rules:

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

• Dependency Pair:

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

Rules:

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

• Dependency Pairs:

DIV(plus(x, y), z) -> DIV(y, z)
DIV(plus(x, y), z) -> DIV(x, z)
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))

Rules:

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pairs:

MINUS(s(x), s(s(s(s(s(s(x''))))))) -> MINUS(p(s(x)), s(s(s(s(s(p(s(x''))))))))
MINUS(s(s(x'')), s(s(s(s(s(x''')))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x''')))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x''')))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(s(x0))))) -> MINUS(s(s(p(s(x'')))), s(s(s(p(s(x0))))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))

Rules:

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

• Dependency Pair:

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

Rules:

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

• Dependency Pairs:

DIV(plus(x, y), z) -> DIV(y, z)
DIV(plus(x, y), z) -> DIV(x, z)
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))

Rules:

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pairs:

MINUS(s(x), s(s(s(s(s(s(x''))))))) -> MINUS(p(s(x)), s(s(s(s(s(p(s(x''))))))))
MINUS(s(s(x'')), s(s(s(s(s(x''')))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x''')))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x''')))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x''')))))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(x''))), s(s(s(s(x0))))) -> MINUS(s(s(p(s(x'')))), s(s(s(p(s(x0))))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(x'')), s(s(s(s(s(x0)))))) -> MINUS(s(p(s(x''))), s(s(s(s(p(s(x0)))))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(x0))), s(s(s(s(x'))))) -> MINUS(s(s(p(s(x0)))), s(s(s(p(s(x'))))))
MINUS(s(s(s(s(x')))), s(s(s(x'')))) -> MINUS(s(s(s(p(s(x'))))), s(s(p(s(x'')))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(x'))), s(s(s(s(x''))))) -> MINUS(s(s(p(s(x')))), s(s(s(p(s(x''))))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(x0)))), s(s(s(x')))) -> MINUS(s(s(s(p(s(x0))))), s(s(p(s(x')))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(x'')))), s(s(s(x0)))) -> MINUS(s(s(s(p(s(x''))))), s(s(p(s(x0)))))
MINUS(s(s(s(s(s(x0))))), s(s(x'))) -> MINUS(s(s(s(s(p(s(x0)))))), s(p(s(x'))))
MINUS(s(s(s(s(s(x'))))), s(s(x''))) -> MINUS(s(s(s(s(p(s(x')))))), s(p(s(x''))))
MINUS(s(s(s(s(s(s(x'')))))), s(y)) -> MINUS(s(s(s(s(s(p(s(x''))))))), p(s(y)))
MINUS(s(s(x''')), s(s(s(s(s(x')))))) -> MINUS(s(p(s(x'''))), s(s(s(s(p(s(x')))))))

Rules:

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

• Dependency Pair:

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

Rules:

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

• Dependency Pairs:

DIV(plus(x, y), z) -> DIV(y, z)
DIV(plus(x, y), z) -> DIV(x, z)
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))

Rules:

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

Termination of R could not be shown.
Duration:
0:30 minutes