Term Rewriting System R:
[X, Z, Y]
h(X, Z) -> f(X, s(X), Z)
f(X, Y, g(X, Y)) -> h(0, g(X, Y))
g(0, Y) -> 0
g(X, s(Y)) -> g(X, Y)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

H(X, Z) -> F(X, s(X), Z)
F(X, Y, g(X, Y)) -> H(0, g(X, Y))
G(X, s(Y)) -> G(X, Y)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Inst


Dependency Pair:

G(X, s(Y)) -> G(X, Y)


Rules:


h(X, Z) -> f(X, s(X), Z)
f(X, Y, g(X, Y)) -> h(0, g(X, Y))
g(0, Y) -> 0
g(X, s(Y)) -> g(X, Y)





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

G(X, s(Y)) -> G(X, Y)
one new Dependency Pair is created:

G(X'', s(s(Y''))) -> G(X'', s(Y''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Inst


Dependency Pair:

G(X'', s(s(Y''))) -> G(X'', s(Y''))


Rules:


h(X, Z) -> f(X, s(X), Z)
f(X, Y, g(X, Y)) -> h(0, g(X, Y))
g(0, Y) -> 0
g(X, s(Y)) -> g(X, Y)





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

G(X'', s(s(Y''))) -> G(X'', s(Y''))
one new Dependency Pair is created:

G(X'''', s(s(s(Y'''')))) -> G(X'''', s(s(Y'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Inst


Dependency Pair:

G(X'''', s(s(s(Y'''')))) -> G(X'''', s(s(Y'''')))


Rules:


h(X, Z) -> f(X, s(X), Z)
f(X, Y, g(X, Y)) -> h(0, g(X, Y))
g(0, Y) -> 0
g(X, s(Y)) -> g(X, Y)





The following dependency pair can be strictly oriented:

G(X'''', s(s(s(Y'''')))) -> G(X'''', s(s(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(s(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


h(X, Z) -> f(X, s(X), Z)
f(X, Y, g(X, Y)) -> h(0, g(X, Y))
g(0, Y) -> 0
g(X, s(Y)) -> g(X, Y)





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Instantiation Transformation


Dependency Pairs:

F(X, Y, g(X, Y)) -> H(0, g(X, Y))
H(X, Z) -> F(X, s(X), Z)


Rules:


h(X, Z) -> f(X, s(X), Z)
f(X, Y, g(X, Y)) -> h(0, g(X, Y))
g(0, Y) -> 0
g(X, s(Y)) -> g(X, Y)





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

H(X, Z) -> F(X, s(X), Z)
one new Dependency Pair is created:

H(0, Z') -> F(0, s(0), Z')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Instantiation Transformation


Dependency Pairs:

H(0, Z') -> F(0, s(0), Z')
F(X, Y, g(X, Y)) -> H(0, g(X, Y))


Rules:


h(X, Z) -> f(X, s(X), Z)
f(X, Y, g(X, Y)) -> h(0, g(X, Y))
g(0, Y) -> 0
g(X, s(Y)) -> g(X, Y)





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

F(X, Y, g(X, Y)) -> H(0, g(X, Y))
one new Dependency Pair is created:

F(0, s(0), g(0, s(0))) -> H(0, g(0, s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Inst
             ...
               →DP Problem 7
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

F(0, s(0), g(0, s(0))) -> H(0, g(0, s(0)))
H(0, Z') -> F(0, s(0), Z')


Rules:


h(X, Z) -> f(X, s(X), Z)
f(X, Y, g(X, Y)) -> h(0, g(X, Y))
g(0, Y) -> 0
g(X, s(Y)) -> g(X, Y)




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