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

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering


Dependency Pairs:

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:


div(0, y) -> 0
div(x, y) -> quot(x, y, y)
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:

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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2, x3) -> x1
s(x1) -> s(x1)
DIV(x1, x2) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳Argument Filtering and Ordering


Dependency Pairs:

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


Rules:


div(0, y) -> 0
div(x, y) -> quot(x, y, y)
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:

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


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{DIV, QUOT}
0 > s

resulting in one new DP problem.
Used Argument Filtering System:
DIV(x1, x2) -> DIV(x1, x2)
QUOT(x1, x2, x3) -> QUOT(x1, x2)
s(x1) -> s


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳AFS
             ...
               →DP Problem 3
                 ↳Dependency Graph


Dependency Pair:

DIV(x, y) -> QUOT(x, y, y)


Rules:


div(0, y) -> 0
div(x, y) -> quot(x, y, y)
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.

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