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:


   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:


   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:

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