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`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳AFS`

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 w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
++ > REV1

resulting in one new DP problem.
Used Argument Filtering System:
REV1(x1, x2) -> REV1(x1, x2)
++(x1, x2) -> ++(x1, x2)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 3`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳AFS`

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`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Argument Filtering and Ordering`

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

The following dependency pairs can be strictly oriented:

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)

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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
++ > REV2 > REV

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`           →DP Problem 4`
`             ↳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