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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
DOUBLE(s(x)) -> DOUBLE(x)
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> MINUS(x, y)
PLUS(s(x), y) -> DOUBLE(y)
PLUS(s(plus(x, y)), z) -> PLUS(plus(x, y), z)

Furthermore, R contains three SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →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)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

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 implicit AFS that need to be oriented.

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

resulting in one new DP problem.

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

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

DOUBLE(s(x)) -> DOUBLE(x)

Rules:

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

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

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

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pairs:

PLUS(s(plus(x, y)), z) -> PLUS(plus(x, y), z)
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)

Rules:

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

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

PLUS(s(plus(x, y)), z) -> PLUS(plus(x, y), z)
five new Dependency Pairs are created:

PLUS(s(plus(0, y'')), z) -> PLUS(y'', z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(plus(x'', s(y'')), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(plus(x'', s(y'')), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(plus(0, y'')), z) -> PLUS(y'', z)
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))

Rules:

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

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

PLUS(s(plus(s(x''), y'')), z) -> PLUS(plus(x'', s(y'')), z)
five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

PLUS(s(plus(s(s(plus(x', y'))), y''')), z) -> PLUS(s(plus(plus(x', y'), s(y'''))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(minus(x', s(y''')), double(s(y''')))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(plus(x', s(s(y'''))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(x', s(y'''))), z)
PLUS(s(plus(s(0), y''')), z) -> PLUS(s(y'''), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(plus(0, y'')), z) -> PLUS(y'', z)
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)

Rules:

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

The following dependency pairs can be strictly oriented:

PLUS(s(plus(s(0), y''')), z) -> PLUS(s(y'''), z)
PLUS(s(plus(0, y'')), z) -> PLUS(y'', z)

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

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

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Narrowing Transformation`

Dependency Pairs:

PLUS(s(plus(s(s(plus(x', y'))), y''')), z) -> PLUS(s(plus(plus(x', y'), s(y'''))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(minus(x', s(y''')), double(s(y''')))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(plus(x', s(s(y'''))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(x', s(y'''))), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)

Rules:

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

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

PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(plus(x', s(s(y'''))), z)
five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

PLUS(s(plus(s(s(s(plus(x'', y')))), y'''')), z) -> PLUS(s(plus(plus(x'', y'), s(s(y'''')))), z)
PLUS(s(plus(s(s(s(x''))), y'''')), z) -> PLUS(s(plus(minus(x'', s(s(y''''))), double(s(s(y''''))))), z)
PLUS(s(plus(s(s(s(x''))), y'''')), z) -> PLUS(plus(x'', s(s(s(y'''')))), z)
PLUS(s(plus(s(s(s(x''))), y'''')), z) -> PLUS(s(plus(x'', s(s(y'''')))), z)
PLUS(s(plus(s(s(0)), y'''')), z) -> PLUS(s(s(y'''')), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(minus(x', s(y''')), double(s(y''')))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(x', s(y'''))), z)
PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(plus(s(s(plus(x', y'))), y''')), z) -> PLUS(s(plus(plus(x', y'), s(y'''))), z)

Rules:

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

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

PLUS(s(plus(s(s(s(x''))), y'''')), z) -> PLUS(plus(x'', s(s(s(y'''')))), z)
five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

PLUS(s(plus(s(s(s(s(plus(x', y'))))), y''''')), z) -> PLUS(s(plus(plus(x', y'), s(s(s(y'''''))))), z)
PLUS(s(plus(s(s(s(s(x')))), y''''')), z) -> PLUS(s(plus(minus(x', s(s(s(y''''')))), double(s(s(s(y''''')))))), z)
PLUS(s(plus(s(s(s(s(x')))), y''''')), z) -> PLUS(plus(x', s(s(s(s(y'''''))))), z)
PLUS(s(plus(s(s(s(s(x')))), y''''')), z) -> PLUS(s(plus(x', s(s(s(y'''''))))), z)
PLUS(s(plus(s(s(s(0))), y''''')), z) -> PLUS(s(s(s(y'''''))), z)
PLUS(s(plus(s(s(s(x''))), y'''')), z) -> PLUS(s(plus(minus(x'', s(s(y''''))), double(s(s(y''''))))), z)
PLUS(s(plus(s(s(s(x''))), y'''')), z) -> PLUS(s(plus(x'', s(s(y'''')))), z)
PLUS(s(plus(s(s(0)), y'''')), z) -> PLUS(s(s(y'''')), z)
PLUS(s(plus(s(s(plus(x', y'))), y''')), z) -> PLUS(s(plus(plus(x', y'), s(y'''))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(minus(x', s(y''')), double(s(y''')))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(x', s(y'''))), z)
PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(plus(s(s(s(plus(x'', y')))), y'''')), z) -> PLUS(s(plus(plus(x'', y'), s(s(y'''')))), z)

Rules:

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

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

PLUS(s(plus(s(s(s(s(x')))), y''''')), z) -> PLUS(plus(x', s(s(s(s(y'''''))))), z)
five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 11`
`                 ↳Polynomial Ordering`

Dependency Pairs:

PLUS(s(plus(s(s(s(s(s(plus(x'', y')))))), y'''''')), z) -> PLUS(s(plus(plus(x'', y'), s(s(s(s(y'''''')))))), z)
PLUS(s(plus(s(s(s(s(s(x''))))), y'''''')), z) -> PLUS(s(plus(minus(x'', s(s(s(s(y''''''))))), double(s(s(s(s(y''''''))))))), z)
PLUS(s(plus(s(s(s(s(s(x''))))), y'''''')), z) -> PLUS(plus(x'', s(s(s(s(s(y'''''')))))), z)
PLUS(s(plus(s(s(s(s(s(x''))))), y'''''')), z) -> PLUS(s(plus(x'', s(s(s(s(y'''''')))))), z)
PLUS(s(plus(s(s(s(s(0)))), y'''''')), z) -> PLUS(s(s(s(s(y'''''')))), z)
PLUS(s(plus(s(s(s(s(x')))), y''''')), z) -> PLUS(s(plus(minus(x', s(s(s(y''''')))), double(s(s(s(y''''')))))), z)
PLUS(s(plus(s(s(s(s(x')))), y''''')), z) -> PLUS(s(plus(x', s(s(s(y'''''))))), z)
PLUS(s(plus(s(s(s(0))), y''''')), z) -> PLUS(s(s(s(y'''''))), z)
PLUS(s(plus(s(s(s(plus(x'', y')))), y'''')), z) -> PLUS(s(plus(plus(x'', y'), s(s(y'''')))), z)
PLUS(s(plus(s(s(s(x''))), y'''')), z) -> PLUS(s(plus(minus(x'', s(s(y''''))), double(s(s(y''''))))), z)
PLUS(s(plus(s(s(s(x''))), y'''')), z) -> PLUS(s(plus(x'', s(s(y'''')))), z)
PLUS(s(plus(s(s(0)), y'''')), z) -> PLUS(s(s(y'''')), z)
PLUS(s(plus(s(s(plus(x', y'))), y''')), z) -> PLUS(s(plus(plus(x', y'), s(y'''))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(minus(x', s(y''')), double(s(y''')))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(x', s(y'''))), z)
PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(plus(s(s(s(s(plus(x', y'))))), y''''')), z) -> PLUS(s(plus(plus(x', y'), s(s(s(y'''''))))), z)

Rules:

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

The following dependency pairs can be strictly oriented:

PLUS(s(plus(s(s(s(s(0)))), y'''''')), z) -> PLUS(s(s(s(s(y'''''')))), z)
PLUS(s(plus(s(s(s(0))), y''''')), z) -> PLUS(s(s(s(y'''''))), z)
PLUS(s(plus(s(s(0)), y'''')), z) -> PLUS(s(s(y'''')), z)

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

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

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 12`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

PLUS(s(plus(s(s(s(s(s(plus(x'', y')))))), y'''''')), z) -> PLUS(s(plus(plus(x'', y'), s(s(s(s(y'''''')))))), z)
PLUS(s(plus(s(s(s(s(s(x''))))), y'''''')), z) -> PLUS(s(plus(minus(x'', s(s(s(s(y''''''))))), double(s(s(s(s(y''''''))))))), z)
PLUS(s(plus(s(s(s(s(s(x''))))), y'''''')), z) -> PLUS(plus(x'', s(s(s(s(s(y'''''')))))), z)
PLUS(s(plus(s(s(s(s(s(x''))))), y'''''')), z) -> PLUS(s(plus(x'', s(s(s(s(y'''''')))))), z)
PLUS(s(plus(s(s(s(s(x')))), y''''')), z) -> PLUS(s(plus(minus(x', s(s(s(y''''')))), double(s(s(s(y''''')))))), z)
PLUS(s(plus(s(s(s(s(x')))), y''''')), z) -> PLUS(s(plus(x', s(s(s(y'''''))))), z)
PLUS(s(plus(s(s(s(plus(x'', y')))), y'''')), z) -> PLUS(s(plus(plus(x'', y'), s(s(y'''')))), z)
PLUS(s(plus(s(s(s(x''))), y'''')), z) -> PLUS(s(plus(minus(x'', s(s(y''''))), double(s(s(y''''))))), z)
PLUS(s(plus(s(s(s(x''))), y'''')), z) -> PLUS(s(plus(x'', s(s(y'''')))), z)
PLUS(s(plus(s(s(plus(x', y'))), y''')), z) -> PLUS(s(plus(plus(x', y'), s(y'''))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(minus(x', s(y''')), double(s(y''')))), z)
PLUS(s(plus(s(s(x')), y''')), z) -> PLUS(s(plus(x', s(y'''))), z)
PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(plus(s(s(s(s(plus(x', y'))))), y''''')), z) -> PLUS(s(plus(plus(x', y'), s(s(s(y'''''))))), z)

Rules:

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

Termination of R could not be shown.
Duration:
0:02 minutes