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
         ↳Argument Filtering and 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)


The following rules can be oriented:

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)


Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
H(x1) -> H(x1)
g(x1, x2) -> g(x1, x2)
f(x1, x2, x3, x4) -> x3
h(x1) -> h(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →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
         ↳AFS
       →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