Term Rewriting System R:
[x]
a(f, a(f, x)) -> a(x, g)
a(x, g) -> a(f, a(g, a(f, x)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

A(f, a(f, x)) -> A(x, g)
A(x, g) -> A(f, a(g, a(f, x)))
A(x, g) -> A(g, a(f, x))
A(x, g) -> A(f, x)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

A(x, g) -> A(f, x)
A(x, g) -> A(f, a(g, a(f, x)))
A(f, a(f, x)) -> A(x, g)


Rules:


a(f, a(f, x)) -> a(x, g)
a(x, g) -> a(f, a(g, a(f, x)))





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

A(x, g) -> A(f, a(g, a(f, x)))
two new Dependency Pairs are created:

A(a(f, x''), g) -> A(f, a(g, a(x'', g)))
A(g, g) -> A(f, a(g, a(f, a(g, a(f, f)))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Semantic Labelling


Dependency Pairs:

A(a(f, x''), g) -> A(f, a(g, a(x'', g)))
A(f, a(f, x)) -> A(x, g)
A(x, g) -> A(f, x)


Rules:


a(f, a(f, x)) -> a(x, g)
a(x, g) -> a(f, a(g, a(f, x)))





Using Semantic Labelling, the DP problem could be transformed. The following model was found:
A(x0, x1)=  1
g=  0
f=  0
a(x0, x1)=  1

From the dependency graph we obtain 2 (labeled) SCCs which each result in correspondingDP problem.


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳SemLab
             ...
               →DP Problem 3
                 ↳Modular Removal of Rules


Dependency Pair:

A00(x, g) -> A00(f, x)


Rules:


a01(f, a00(f, x)) -> a00(x, g)
a01(f, a01(f, x)) -> a10(x, g)
a00(x, g) -> a01(f, a01(g, a00(f, x)))
a10(x, g) -> a01(f, a01(g, a01(f, x)))





We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(A_00(x1, x2))=  x1 + x2  
  POL(g)=  0  
  POL(f)=  0  

We have the following set D of usable symbols: {A00, f}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

A00(x, g) -> A00(f, x)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳SemLab
             ...
               →DP Problem 4
                 ↳Modular Removal of Rules


Dependency Pairs:

A10(x, g) -> A01(f, x)
A01(f, a01(f, x)) -> A10(x, g)


Rules:


a01(f, a00(f, x)) -> a00(x, g)
a01(f, a01(f, x)) -> a10(x, g)
a00(x, g) -> a01(f, a01(g, a00(f, x)))
a10(x, g) -> a01(f, a01(g, a01(f, x)))





We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(A_01(x1, x2))=  x1 + x2  
  POL(a_01(x1, x2))=  x1 + x2  
  POL(g)=  0  
  POL(A_10(x1, x2))=  x1 + x2  
  POL(f)=  0  

We have the following set D of usable symbols: {A01, g, A10, f}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

A01(f, a01(f, x)) -> A10(x, g)

4 non usable rules have been deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.


Termination of R successfully shown.
Duration:
0:02 minutes