Termination Proof using AProVE

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.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
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 AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and 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 AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

DOUBLE(x₁) -> DOUBLE(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

Dependency 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)

The following usable rules w.r.t. to the 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: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

PLUS(x₁, x₂) -> PLUS(x₁, x₂)
s(x₁) -> x₁
plus(x₁, x₂) -> plus(x₁, x₂)
minus(x₁, x₂) -> x₁
double(x₁) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pairs:

PLUS(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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳Narrowing Transformation

Dependency Pairs:

PLUS(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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳Narrowing Transformation

Dependency Pairs:

PLUS(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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 11

                 ↳Argument Filtering and 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)

The following usable rules w.r.t. to the 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: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

PLUS(x₁, x₂) -> PLUS(x₁, x₂)
s(x₁) -> x₁
plus(x₁, x₂) -> plus(x₁, x₂)
minus(x₁, x₂) -> x₁
double(x₁) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →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