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))))

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`
`         ↳Polynomial 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))))

The following dependency pair can be strictly oriented:

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

There are no usable rules 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`
`         ↳Polo`
`           →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))))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →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))))

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`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Narrowing 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))))

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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 5`
`                 ↳Polynomial Ordering`

Dependency Pairs:

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

The following dependency pairs can be strictly oriented:

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

Additionally, the following usable rules 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`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 6`
`                 ↳Dependency Graph`

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes