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`
`                 ↳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.

`   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`
`                 ↳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.

`   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