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
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →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 using the Ce-refinement 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.
Used Argument Filtering System:
P(x1) -> P(x1)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →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
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →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))))





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)


The following usable rules using the Ce-refinement can be oriented:

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))


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

resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
plus(x1, x2) -> plus(x1, x2)
minus(x1, x2) -> minus(x1, x2)
s(x1) -> s(x1)
p(x1) -> p(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →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:
0:00 minutes