Term Rewriting System R:
[x, y, z]
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

PLUS(s(x), y) -> PLUS(x, y)
TIMES(s(x), y) -> PLUS(y, times(x, y))
TIMES(s(x), y) -> TIMES(x, y)
DIV(x, y) -> QUOT(x, y, y)
DIV(div(x, y), z) -> DIV(x, times(y, z))
DIV(div(x, y), z) -> TIMES(y, z)
QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> DIV(x, s(z))

Furthermore, R contains three SCCs. 


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


Dependency Pair:

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


Rules:


plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(PLUS(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
         ↳Remaining


Dependency Pair:


Rules:


plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(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
         ↳Remaining


Dependency Pair:

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


Rules:


plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(TIMES(x1, x2))=  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
         ↳Remaining


Dependency Pair:


Rules:


plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))





Using the Dependency Graph resulted in no new DP problems.


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




The following remains to be proven:
Dependency Pairs:

DIV(div(x, y), z) -> DIV(x, times(y, z))
QUOT(x, 0, s(z)) -> DIV(x, s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)
DIV(x, y) -> QUOT(x, y, y)


Rules:


plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))




Termination of R could not be shown.
Duration:
0:00 minutes