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
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


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))


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

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
++' > ++

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


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
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS


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


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳AFS


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
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Argument Filtering and 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)


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
           →DP Problem 6
             ↳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