Term Rewriting System R:
[x, y, z]
rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

REV(++(x, y)) -> REV1(x, y)
REV(++(x, y)) -> REV2(x, y)
REV1(x, ++(y, z)) -> REV1(y, z)
REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) -> REV(rev2(y, z))
REV2(x, ++(y, z)) -> REV2(y, z)

Furthermore, R contains two SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
Nar


Dependency Pair:

REV1(x, ++(y, z)) -> REV1(y, z)


Rules:


rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

REV1(x, ++(y, z)) -> REV1(y, z)


There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(REV1(x1, x2))=  x1 + x2  
  POL(++(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
REV1(x1, x2) -> REV1(x1, x2)
++(x1, x2) -> ++(x1, x2)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 3
Dependency Graph
       →DP Problem 2
Nar


Dependency Pair:


Rules:


rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Narrowing Transformation


Dependency Pairs:

REV2(x, ++(y, z)) -> REV2(y, z)
REV2(x, ++(y, z)) -> REV(rev2(y, z))
REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
REV(++(x, y)) -> REV2(x, y)


Rules:


rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

REV2(x, ++(y, z)) -> REV(rev2(y, z))
two new Dependency Pairs are created:

REV2(x, ++(y', nil)) -> REV(nil)
REV2(x, ++(y0, ++(y'', z''))) -> REV(rev(++(y0, rev(rev2(y'', z'')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 4
Rewriting Transformation


Dependency Pairs:

REV2(x, ++(y0, ++(y'', z''))) -> REV(rev(++(y0, rev(rev2(y'', z'')))))
REV(++(x, y)) -> REV2(x, y)
REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) -> REV2(y, z)


Rules:


rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

REV2(x, ++(y0, ++(y'', z''))) -> REV(rev(++(y0, rev(rev2(y'', z'')))))
one new Dependency Pair is created:

REV2(x, ++(y0, ++(y'', z''))) -> REV(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 4
Rw
             ...
               →DP Problem 5
Forward Instantiation Transformation


Dependency Pairs:

REV2(x, ++(y0, ++(y'', z''))) -> REV(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))
REV2(x, ++(y, z)) -> REV2(y, z)
REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
REV(++(x, y)) -> REV2(x, y)


Rules:


rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

REV(++(x, y)) -> REV2(x, y)
two new Dependency Pairs are created:

REV(++(x'', ++(y'', z''))) -> REV2(x'', ++(y'', z''))
REV(++(x'', ++(y0'', ++(y'''', z'''')))) -> REV2(x'', ++(y0'', ++(y'''', z'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 4
Rw
             ...
               →DP Problem 6
Narrowing Transformation


Dependency Pairs:

REV2(x, ++(y, z)) -> REV2(y, z)
REV(++(x'', ++(y0'', ++(y'''', z'''')))) -> REV2(x'', ++(y0'', ++(y'''', z'''')))
REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
REV(++(x'', ++(y'', z''))) -> REV2(x'', ++(y'', z''))
REV2(x, ++(y0, ++(y'', z''))) -> REV(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))


Rules:


rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
two new Dependency Pairs are created:

REV2(x, ++(y', nil)) -> REV(++(x, rev(nil)))
REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, rev(rev(++(y0, rev(rev2(y'', z'')))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 4
Rw
             ...
               →DP Problem 7
Rewriting Transformation


Dependency Pairs:

REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, rev(rev(++(y0, rev(rev2(y'', z'')))))))
REV(++(x'', ++(y0'', ++(y'''', z'''')))) -> REV2(x'', ++(y0'', ++(y'''', z'''')))
REV2(x, ++(y', nil)) -> REV(++(x, rev(nil)))
REV(++(x'', ++(y'', z''))) -> REV2(x'', ++(y'', z''))
REV2(x, ++(y0, ++(y'', z''))) -> REV(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))
REV2(x, ++(y, z)) -> REV2(y, z)


Rules:


rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

REV2(x, ++(y', nil)) -> REV(++(x, rev(nil)))
one new Dependency Pair is created:

REV2(x, ++(y', nil)) -> REV(++(x, nil))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 4
Rw
             ...
               →DP Problem 8
Rewriting Transformation


Dependency Pairs:

REV(++(x'', ++(y0'', ++(y'''', z'''')))) -> REV2(x'', ++(y0'', ++(y'''', z'''')))
REV2(x, ++(y0, ++(y'', z''))) -> REV(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))
REV2(x, ++(y, z)) -> REV2(y, z)
REV(++(x'', ++(y'', z''))) -> REV2(x'', ++(y'', z''))
REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, rev(rev(++(y0, rev(rev2(y'', z'')))))))


Rules:


rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, rev(rev(++(y0, rev(rev2(y'', z'')))))))
one new Dependency Pair is created:

REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, rev(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 4
Rw
             ...
               →DP Problem 9
Forward Instantiation Transformation


Dependency Pairs:

REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, rev(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))))
REV(++(x'', ++(y'', z''))) -> REV2(x'', ++(y'', z''))
REV2(x, ++(y0, ++(y'', z''))) -> REV(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))
REV2(x, ++(y, z)) -> REV2(y, z)
REV(++(x'', ++(y0'', ++(y'''', z'''')))) -> REV2(x'', ++(y0'', ++(y'''', z'''')))


Rules:


rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

REV2(x, ++(y, z)) -> REV2(y, z)
two new Dependency Pairs are created:

REV2(x, ++(y0, ++(y'', z''))) -> REV2(y0, ++(y'', z''))
REV2(x, ++(y', ++(y0'', ++(y'''', z'''')))) -> REV2(y', ++(y0'', ++(y'''', z'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 4
Rw
             ...
               →DP Problem 10
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

REV2(x, ++(y', ++(y0'', ++(y'''', z'''')))) -> REV2(y', ++(y0'', ++(y'''', z'''')))
REV2(x, ++(y0, ++(y'', z''))) -> REV2(y0, ++(y'', z''))
REV(++(x'', ++(y0'', ++(y'''', z'''')))) -> REV2(x'', ++(y0'', ++(y'''', z'''')))
REV2(x, ++(y0, ++(y'', z''))) -> REV(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))
REV(++(x'', ++(y'', z''))) -> REV2(x'', ++(y'', z''))
REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, rev(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))))


Rules:


rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:13 minutes