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
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst


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)


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))=  x1  
  POL(e(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst


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
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering
       →DP Problem 3
         ↳Inst


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 pair can be strictly oriented:

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


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(c(x1))=  1 + x1  
  POL(0)=  0  
  POL(g(x1, x2))=  0  
  POL(e(x1))=  0  
  POL(D(x1, x2))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Polynomial Ordering
       →DP Problem 3
         ↳Inst


Dependency Pairs:

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 pair can be strictly oriented:

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


Additionally, the following usable rule w.r.t. to the implicit AFS can be oriented:

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


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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Polo
             ...
               →DP Problem 6
                 ↳Polynomial Ordering
       →DP Problem 3
         ↳Inst


Dependency Pair:

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 pair can be strictly oriented:

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


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 + x2  
  POL(D(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Polo
             ...
               →DP Problem 7
                 ↳Dependency Graph
       →DP Problem 3
         ↳Inst


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
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Instantiation Transformation


Dependency Pair:

H(z, e(x)) -> H(c(z), d(z, x))


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))





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

H(z, e(x)) -> H(c(z), d(z, x))
one new Dependency Pair is created:

H(c(z''), e(x')) -> H(c(c(z'')), d(c(z''), x'))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
           →DP Problem 8
             ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

H(c(z''), e(x')) -> H(c(c(z'')), d(c(z''), x'))


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))




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