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

         ↳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)

     ↳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.

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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

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