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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

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


Rules:


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





The following dependency pairs can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(++'(x1, x2))=  x2  
  POL(++(x1, x2))=  1 + x1 + x2  
  POL(nil)=  0  
  POL(.(x1, x2))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pair:

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


Rules:


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





The following dependency pair can be strictly oriented:

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


There are no usable rules 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 3
             ↳Polo
             ...
               →DP Problem 4
                 ↳Dependency Graph
       →DP Problem 2
         ↳Polo


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

REV(++(x, y)) -> REV(x)
REV(++(x, y)) -> REV(y)


Rules:


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





The following dependency pairs can be strictly oriented:

REV(++(x, y)) -> REV(x)
REV(++(x, y)) -> REV(y)


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

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.

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