Term Rewriting System R:
[x, y, z]
f(0, 1, g(x, y), z) -> f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) -> 0
g(0, 1) -> 1
h(g(x, y)) -> h(x)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(0, 1, g(x, y), z) -> F(g(x, y), g(x, y), g(x, y), h(x))
F(0, 1, g(x, y), z) -> H(x)
H(g(x, y)) -> H(x)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

H(g(x, y)) -> H(x)


Rules:


f(0, 1, g(x, y), z) -> f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) -> 0
g(0, 1) -> 1
h(g(x, y)) -> h(x)





The following dependency pair can be strictly oriented:

H(g(x, y)) -> H(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(g(x1, x2))=  1 + x1  
  POL(H(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Remaining


Dependency Pair:


Rules:


f(0, 1, g(x, y), z) -> f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) -> 0
g(0, 1) -> 1
h(g(x, y)) -> h(x)





Using the Dependency Graph resulted in no new DP problems.


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




The following remains to be proven:
Dependency Pair:

F(0, 1, g(x, y), z) -> F(g(x, y), g(x, y), g(x, y), h(x))


Rules:


f(0, 1, g(x, y), z) -> f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) -> 0
g(0, 1) -> 1
h(g(x, y)) -> h(x)




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