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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
+'(f(x), f(y)) -> +'(x, y)
+'(f(x), +(f(y), z)) -> +'(f(+(x, y)), z)
+'(f(x), +(f(y), z)) -> +'(x, y)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering


Dependency Pairs:

+'(f(x), +(f(y), z)) -> +'(x, y)
+'(f(x), +(f(y), z)) -> +'(f(+(x, y)), z)
+'(f(x), f(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))


Rules:


+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)





The following dependency pairs can be strictly oriented:

+'(f(x), +(f(y), z)) -> +'(x, y)
+'(f(x), +(f(y), z)) -> +'(f(+(x, y)), z)
+'(f(x), f(y)) -> +'(x, y)


The following rules can be oriented:

+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(+(x1, x2))=  x1 + x2  
  POL(f(x1))=  1 + x1  
  POL(+'(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> +'(x1, x2)
+(x1, x2) -> +(x1, x2)
f(x1) -> f(x1)


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


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))


Rules:


+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)





The following dependency pair can be strictly oriented:

+'(+(x, y), z) -> +'(y, z)


The following rules can be oriented:

+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(+(x1, x2))=  1 + x1 + x2  
  POL(f(x1))=  x1  
  POL(+'(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> +'(x1, x2)
+(x1, x2) -> +(x1, x2)
f(x1) -> f(x1)


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


Dependency Pair:

+'(+(x, y), z) -> +'(x, +(y, z))


Rules:


+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)





The following dependency pair can be strictly oriented:

+'(+(x, y), z) -> +'(x, +(y, z))


The following rules can be oriented:

+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(+(x1, x2))=  1 + x1 + x2  
  POL(f(x1))=  x1  

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> x1
+(x1, x2) -> +(x1, x2)
f(x1) -> f(x1)


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


Dependency Pair:


Rules:


+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes