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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

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


Rules:


++(nil, y) -> y
++(x, nil) -> x
++(.(x, y), z) -> .(x, ++(y, z))
++(++(x, y), z) -> ++(x, ++(y, z))





The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules using the Ce-refinement can be oriented:

++(nil, y) -> y
++(x, nil) -> x
++(.(x, y), z) -> .(x, ++(y, z))
++(++(x, y), z) -> ++(x, ++(y, z))


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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polynomial Ordering


Dependency Pair:

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


Rules:


++(nil, y) -> y
++(x, nil) -> x
++(.(x, y), z) -> .(x, ++(y, z))
++(++(x, y), z) -> ++(x, ++(y, z))





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement that need to be oriented.

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

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


++(nil, y) -> y
++(x, nil) -> x
++(.(x, y), z) -> .(x, ++(y, z))
++(++(x, y), z) -> ++(x, ++(y, z))





Using the Dependency Graph resulted in no new DP problems.

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