Term Rewriting System R:
[z, x, y]
h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

H(z, e(x)) -> H(c(z), d(z, x))
H(z, e(x)) -> D(z, x)
D(z, g(x, y)) -> G(e(x), d(z, y))
D(z, g(x, y)) -> D(z, y)
D(c(z), g(g(x, y), 0)) -> G(d(c(z), g(x, y)), d(z, g(x, y)))
D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))
D(c(z), g(g(x, y), 0)) -> D(z, g(x, y))
G(e(x), e(y)) -> G(x, y)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Remaining


Dependency Pair:

G(e(x), e(y)) -> G(x, y)


Rules:


h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))





The following dependency pair can be strictly oriented:

G(e(x), e(y)) -> G(x, y)


The following rules can be oriented:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1))=  x1  
  POL(0)=  0  
  POL(G(x1, x2))=  x1 + x2  
  POL(e(x1))=  1 + x1  
  POL(d)=  1  
  POL(h(x1, x2))=  x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
G(x1, x2) -> G(x1, x2)
e(x1) -> e(x1)
h(x1, x2) -> h(x1, x2)
c(x1) -> c(x1)
d(x1, x2) -> d
g(x1, x2) -> x2


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Remaining


Dependency Pair:


Rules:


h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳Remaining


Dependency Pairs:

D(c(z), g(g(x, y), 0)) -> D(z, g(x, y))
D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))
D(z, g(x, y)) -> D(z, y)


Rules:


h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))





The following dependency pairs can be strictly oriented:

D(c(z), g(g(x, y), 0)) -> D(z, g(x, y))
D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))


The following rules can be oriented:

g(e(x), e(y)) -> e(g(x, y))
h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1))=  x1  
  POL(0)=  1  
  POL(g(x1, x2))=  x1 + x2  
  POL(e)=  0  
  POL(d)=  0  
  POL(D(x1, x2))=  1 + x1 + x2  
  POL(h(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
D(x1, x2) -> D(x1, x2)
c(x1) -> c(x1)
g(x1, x2) -> g(x1, x2)
e(x1) -> e
h(x1, x2) -> h(x1, x2)
d(x1, x2) -> d


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




The following remains to be proven:


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




The following remains to be proven:

Termination of R could not be shown.
Duration:
0:02 minutes