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

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

FLATTEN(unit(x)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> ++'(flatten(x), flatten(y))
FLATTEN(++(x, y)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> FLATTEN(y)
FLATTEN(++(unit(x), y)) -> ++'(flatten(x), flatten(y))
FLATTEN(++(unit(x), y)) -> FLATTEN(x)
FLATTEN(++(unit(x), y)) -> FLATTEN(y)
REV(++(x, y)) -> ++'(rev(y), rev(x))
REV(++(x, y)) -> REV(y)
REV(++(x, y)) -> REV(x)
++'(++(x, y), z) -> ++'(x, ++(y, z))
++'(++(x, y), z) -> ++'(y, z)

Furthermore, R contains three SCCs.


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


Dependency Pairs:

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


Rules:


flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(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 rules can be oriented:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x


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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 4
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo


Dependency Pair:


Rules:


flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

FLATTEN(++(unit(x), y)) -> FLATTEN(y)
FLATTEN(++(unit(x), y)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> FLATTEN(y)
FLATTEN(++(x, y)) -> FLATTEN(x)
FLATTEN(unit(x)) -> FLATTEN(x)


Rules:


flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))





The following dependency pairs can be strictly oriented:

FLATTEN(++(unit(x), y)) -> FLATTEN(y)
FLATTEN(++(unit(x), y)) -> FLATTEN(x)
FLATTEN(unit(x)) -> FLATTEN(x)


Additionally, the following rules can be oriented:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(unit(x1))=  1 + x1  
  POL(rev(x1))=  x1  
  POL(flatten(x1))=  0  
  POL(++(x1, x2))=  x1 + x2  
  POL(FLATTEN(x1))=  1 + x1  
  POL(nil)=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 5
Polynomial Ordering
       →DP Problem 3
Polo


Dependency Pairs:

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


Rules:


flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))





The following dependency pairs can be strictly oriented:

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


Additionally, the following rules can be oriented:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(unit(x1))=  x1  
  POL(rev(x1))=  x1  
  POL(flatten(x1))=  x1  
  POL(++(x1, x2))=  1 + x1 + x2  
  POL(FLATTEN(x1))=  1 + x1  
  POL(nil)=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 5
Polo
             ...
               →DP Problem 6
Dependency Graph
       →DP Problem 3
Polo


Dependency Pair:


Rules:


flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

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


Rules:


flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))





The following dependency pairs can be strictly oriented:

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


Additionally, the following rules can be oriented:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(unit(x1))=  x1  
  POL(rev(x1))=  x1  
  POL(REV(x1))=  1 + x1  
  POL(flatten(x1))=  x1  
  POL(++(x1, x2))=  1 + x1 + x2  
  POL(nil)=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 7
Dependency Graph


Dependency Pair:


Rules:


flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(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