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