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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)


Rules:


quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))





The following dependency pair can be strictly oriented:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(QUOT(x1, x2, x3))=  x1  
  POL(0)=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Dependency Graph


Dependency Pair:

QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))


Rules:


quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes