Termination Proof using AProVE

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.

     ↳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.

     ↳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:

     ↳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:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Polynomial 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 implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(REV1(x₁, x₂)) = x₂

POL(++(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳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.

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 15

                 ↳Polynomial 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''''''))))

Additionally, the following usable rules for innermost w.r.t. to the implicit 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: Polynomial ordering with Polynomial interpretation:

POL(rev2(x₁, x₂)) = x₂

POL(rev(x₁)) = x₁

POL(REV(x₁)) = x₁

POL(rev1(x₁, x₂)) = 0

POL(++(x₁, x₂)) = 1 + x₂

POL(REV2(x₁, x₂)) = x₂

POL(nil) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 16

                 ↳Dependency Graph

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.

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

POL(REV1(x₁, x₂))	= x₂
POL(++(x₁, x₂))	= 1 + x₂

POL(rev2(x₁, x₂))	= x₂
POL(rev(x₁))	= x₁
POL(REV(x₁))	= x₁
POL(rev1(x₁, x₂))	= 0
POL(++(x₁, x₂))	= 1 + x₂
POL(REV2(x₁, x₂))	= x₂
POL(nil)	= 0