Term Rewriting System R:
[x, y, z]
f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(0, 1, x) -> F(g(x), g(x), x)
F(g(x), y, z) -> F(x, y, z)
F(x, g(y), z) -> F(x, y, z)
F(x, y, g(z)) -> F(x, y, z)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Negative Polynomial Order


Dependency Pairs:

F(x, y, g(z)) -> F(x, y, z)
F(x, g(y), z) -> F(x, y, z)
F(g(x), y, z) -> F(x, y, z)
F(0, 1, x) -> F(g(x), g(x), x)


Rules:


f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))





The following Dependency Pair can be strictly oriented using the given order.

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


There are no usable rules (regarding the implicit AFS).
Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, ..., x3) ) = x3

POL( g(x1) ) = x1 + 1


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳Neg POLO
           →DP Problem 2
             ↳Semantic Labelling


Dependency Pairs:

F(x, g(y), z) -> F(x, y, z)
F(g(x), y, z) -> F(x, y, z)
F(0, 1, x) -> F(g(x), g(x), x)


Rules:


f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))





Using Semantic Labelling, the DP problem could be transformed. The following model was found:
f(x0, x1, x2)=  1
g(x0)=  x0
F(x0, x1, x2)=  1 + x2
0=  0
1=  1

From the dependency graph we obtain 8 (labeled) SCCs which each result in correspondingDP problem.


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


Dependency Pairs:

F111(g1(x), y, z) -> F111(x, y, z)
F111(x, g1(y), z) -> F111(x, y, z)


Rules:


f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))





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(F_111(x1, x2, x3))=  x1 + x2 + x3  
  POL(g_1(x1))=  x1  

We have the following set D of usable symbols: {F111}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F111(g1(x), y, z) -> F111(x, y, z)
F111(x, g1(y), z) -> F111(x, y, z)

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
         ↳Neg POLO
           →DP Problem 2
             ↳SemLab
             ...
               →DP Problem 4
                 ↳Modular Removal of Rules


Dependency Pairs:

F100(g1(x), y, z) -> F100(x, y, z)
F100(x, g0(y), z) -> F100(x, y, z)


Rules:


f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))





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(F_100(x1, x2, x3))=  x1 + x2 + x3  
  POL(g_1(x1))=  x1  
  POL(g_0(x1))=  x1  

We have the following set D of usable symbols: {F100}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F100(g1(x), y, z) -> F100(x, y, z)
F100(x, g0(y), z) -> F100(x, y, z)

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
         ↳Neg POLO
           →DP Problem 2
             ↳SemLab
             ...
               →DP Problem 5
                 ↳Modular Removal of Rules


Dependency Pairs:

F011(g0(x), y, z) -> F011(x, y, z)
F011(x, g1(y), z) -> F011(x, y, z)


Rules:


f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))





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(g_1(x1))=  x1  
  POL(F_011(x1, x2, x3))=  x1 + x2 + x3  
  POL(g_0(x1))=  x1  

We have the following set D of usable symbols: {F011}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F011(g0(x), y, z) -> F011(x, y, z)
F011(x, g1(y), z) -> F011(x, y, z)

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
         ↳Neg POLO
           →DP Problem 2
             ↳SemLab
             ...
               →DP Problem 6
                 ↳Modular Removal of Rules


Dependency Pairs:

F110(g1(x), y, z) -> F110(x, y, z)
F110(x, g1(y), z) -> F110(x, y, z)


Rules:


f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))





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(g_1(x1))=  x1  
  POL(F_110(x1, x2, x3))=  x1 + x2 + x3  

We have the following set D of usable symbols: {F110}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F110(g1(x), y, z) -> F110(x, y, z)
F110(x, g1(y), z) -> F110(x, y, z)

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
         ↳Neg POLO
           →DP Problem 2
             ↳SemLab
             ...
               →DP Problem 7
                 ↳Modular Removal of Rules


Dependency Pairs:

F001(g0(x), y, z) -> F001(x, y, z)
F001(x, g0(y), z) -> F001(x, y, z)


Rules:


f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))





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(F_001(x1, x2, x3))=  x1 + x2 + x3  
  POL(g_0(x1))=  x1  

We have the following set D of usable symbols: {F001}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F001(g0(x), y, z) -> F001(x, y, z)
F001(x, g0(y), z) -> F001(x, y, z)

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
         ↳Neg POLO
           →DP Problem 2
             ↳SemLab
             ...
               →DP Problem 8
                 ↳Modular Removal of Rules


Dependency Pairs:

F000(g0(x), y, z) -> F000(x, y, z)
F000(x, g0(y), z) -> F000(x, y, z)


Rules:


f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))





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(F_000(x1, x2, x3))=  x1 + x2 + x3  
  POL(g_0(x1))=  x1  

We have the following set D of usable symbols: {F000}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F000(g0(x), y, z) -> F000(x, y, z)
F000(x, g0(y), z) -> F000(x, y, z)

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
         ↳Neg POLO
           →DP Problem 2
             ↳SemLab
             ...
               →DP Problem 9
                 ↳Modular Removal of Rules


Dependency Pairs:

F101(g1(x), y, z) -> F101(x, y, z)
F101(x, g0(y), z) -> F101(x, y, z)


Rules:


f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))





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(g_1(x1))=  x1  
  POL(F_101(x1, x2, x3))=  x1 + x2 + x3  
  POL(g_0(x1))=  x1  

We have the following set D of usable symbols: {F101}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F101(g1(x), y, z) -> F101(x, y, z)
F101(x, g0(y), z) -> F101(x, y, z)

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
         ↳Neg POLO
           →DP Problem 2
             ↳SemLab
             ...
               →DP Problem 10
                 ↳Modular Removal of Rules


Dependency Pairs:

F010(g0(x), y, z) -> F010(x, y, z)
F010(x, g1(y), z) -> F010(x, y, z)


Rules:


f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))





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(F_010(x1, x2, x3))=  x1 + x2 + x3  
  POL(g_1(x1))=  x1  
  POL(g_0(x1))=  x1  

We have the following set D of usable symbols: {F010}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F010(g0(x), y, z) -> F010(x, y, z)
F010(x, g1(y), z) -> F010(x, y, z)

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.


Termination of R successfully shown.
Duration:
0:07 minutes