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))

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


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the implicit AFS 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))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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