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
         ↳Forward Instantiation Transformation
       →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




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

REV1(x, ++(y, z)) -> REV1(y, z)
one new Dependency Pair is created:

REV1(x, ++(y0, ++(y'', z''))) -> REV1(y0, ++(y'', z''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Nar


Dependency Pair:

REV1(x, ++(y0, ++(y'', z''))) -> REV1(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

REV1(x, ++(y0, ++(y'', z''))) -> REV1(y0, ++(y'', z''))
one new Dependency Pair is created:

REV1(x, ++(y0'', ++(y''0, ++(y'''', z'''')))) -> REV1(y0'', ++(y''0, ++(y'''', z'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Nar


Dependency Pair:

REV1(x, ++(y0'', ++(y''0, ++(y'''', z'''')))) -> REV1(y0'', ++(y''0, ++(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, ++(y0'', ++(y''0, ++(y'''', z'''')))) -> REV1(y0'', ++(y''0, ++(y'''', z'''')))


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
REV1(x1, x2) -> x2
++(x1, x2) -> ++(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳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
         ↳FwdInst
       →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
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳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
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Rw
             ...
               →DP Problem 7
                 ↳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
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Rw
             ...
               →DP Problem 8
                 ↳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
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Rw
             ...
               →DP Problem 9
                 ↳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
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Rw
             ...
               →DP Problem 10
                 ↳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
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Rw
             ...
               →DP Problem 11
                 ↳Rewriting 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 Rewriting SCC transformation can be performed.
As a result of transforming the rule

REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, rev(++(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))))))
one new Dependency Pair is created:

REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, ++(rev1(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))), rev2(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
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Rw
             ...
               →DP Problem 12
                 ↳Forward Instantiation Transformation


Dependency Pairs:

REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, ++(rev1(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))), rev2(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(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'')))))
REV2(x, ++(y, z)) -> REV2(y, z)
REV(++(x'', ++(y'', z''))) -> REV2(x'', ++(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
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Rw
             ...
               →DP Problem 13
                 ↳Forward Instantiation Transformation


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, ++(rev1(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))), rev2(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 Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

REV(++(x''', ++(y''0, ++(y'''', z'''')))) -> REV2(x''', ++(y''0, ++(y'''', z'''')))
REV(++(x''', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))) -> REV2(x''', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Rw
             ...
               →DP Problem 14
                 ↳Forward Instantiation Transformation


Dependency Pairs:

REV(++(x''', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))) -> REV2(x''', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))
REV2(x, ++(y0, ++(y'', z''))) -> REV2(y0, ++(y'', z''))
REV(++(x''', ++(y''0, ++(y'''', z'''')))) -> REV2(x''', ++(y''0, ++(y'''', z'''')))
REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, ++(rev1(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))), rev2(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(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'')))))
REV2(x, ++(y', ++(y0'', ++(y'''', z'''')))) -> REV2(y', ++(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, ++(y0, ++(y'', z''))) -> REV2(y0, ++(y'', z''))
two new Dependency Pairs are created:

REV2(x, ++(y0'', ++(y''0, ++(y'''', z'''')))) -> REV2(y0'', ++(y''0, ++(y'''', z'''')))
REV2(x, ++(y0', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))) -> REV2(y0', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Rw
             ...
               →DP Problem 15
                 ↳Argument Filtering and Ordering


Dependency Pairs:

REV2(x, ++(y0', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))) -> REV2(y0', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))
REV2(x, ++(y0'', ++(y''0, ++(y'''', z'''')))) -> REV2(y0'', ++(y''0, ++(y'''', z'''')))
REV2(x, ++(y', ++(y0'', ++(y'''', z'''')))) -> REV2(y', ++(y0'', ++(y'''', z'''')))
REV(++(x''', ++(y''0, ++(y'''', z'''')))) -> REV2(x''', ++(y''0, ++(y'''', z'''')))
REV2(x, ++(y0, ++(y'', z''))) -> REV(++(x, ++(rev1(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(y'', z'')))), rev2(rev1(y0, rev(rev2(y'', z''))), rev2(y0, rev(rev2(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'''', ++(y0'''', ++(y'''''', z''''''))))) -> REV2(x''', ++(y'''', ++(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




The following dependency pairs can be strictly oriented:

REV2(x, ++(y0', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))) -> REV2(y0', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))
REV2(x, ++(y0'', ++(y''0, ++(y'''', z'''')))) -> REV2(y0'', ++(y''0, ++(y'''', z'''')))
REV2(x, ++(y', ++(y0'', ++(y'''', z'''')))) -> REV2(y', ++(y0'', ++(y'''', z'''')))
REV(++(x''', ++(y''0, ++(y'''', z'''')))) -> REV2(x''', ++(y''0, ++(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'''', ++(y0'''', ++(y'''''', z''''''))))) -> REV2(x''', ++(y'''', ++(y0'''', ++(y'''''', z''''''))))


The following usable rules for innermost w.r.t. to the AFS can be oriented:

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


Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
REV2(x1, x2) -> x2
++(x1, x2) -> ++(x2)
REV(x1) -> x1
rev2(x1, x2) -> x2
rev(x1) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Rw
             ...
               →DP Problem 16
                 ↳Dependency Graph


Dependency Pair:

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




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes