Term Rewriting System R:
[x, l, y]
rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

REV(cons(x, l)) -> REV1(x, l)
REV(cons(x, l)) -> REV2(x, l)
REV1(x, cons(y, l)) -> REV1(y, l)
REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))
REV2(x, cons(y, l)) -> REV2(y, l)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

REV1(x, cons(y, l)) -> REV1(y, l)

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

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

REV1(x, cons(y, l)) -> REV1(y, l)
one new Dependency Pair is created:

REV1(x, cons(y0, cons(y'', l''))) -> REV1(y0, cons(y'', l''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

REV1(x, cons(y0, cons(y'', l''))) -> REV1(y0, cons(y'', l''))

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

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

REV1(x, cons(y0, cons(y'', l''))) -> REV1(y0, cons(y'', l''))
one new Dependency Pair is created:

REV1(x, cons(y0'', cons(y''0, cons(y'''', l'''')))) -> REV1(y0'', cons(y''0, cons(y'''', l'''')))

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`
`         ↳FwdInst`

Dependency Pair:

REV1(x, cons(y0'', cons(y''0, cons(y'''', l'''')))) -> REV1(y0'', cons(y''0, cons(y'''', l'''')))

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

REV1(x, cons(y0'', cons(y''0, cons(y'''', l'''')))) -> REV1(y0'', cons(y''0, cons(y'''', l'''')))

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(cons(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`
`         ↳FwdInst`

Dependency Pair:

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Forward Instantiation Transformation`

Dependency Pairs:

REV2(x, cons(y, l)) -> REV2(y, l)
REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))
REV(cons(x, l)) -> REV2(x, l)

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

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

REV(cons(x, l)) -> REV2(x, l)
one new Dependency Pair is created:

REV(cons(x'', cons(y'', l''))) -> REV2(x'', cons(y'', l''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Narrowing Transformation`

Dependency Pairs:

REV(cons(x'', cons(y'', l''))) -> REV2(x'', cons(y'', l''))
REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))
REV2(x, cons(y, l)) -> REV2(y, l)

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

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

REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))
two new Dependency Pairs are created:

REV2(x, cons(y', nil)) -> REV(cons(x, nil))
REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, rev(cons(y0, rev2(y'', l'')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Rewriting Transformation`

Dependency Pairs:

REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, rev(cons(y0, rev2(y'', l'')))))
REV2(x, cons(y, l)) -> REV2(y, l)
REV(cons(x'', cons(y'', l''))) -> REV2(x'', cons(y'', l''))

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

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

REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, rev(cons(y0, rev2(y'', l'')))))
one new Dependency Pair is created:

REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, cons(rev1(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

REV(cons(x'', cons(y'', l''))) -> REV2(x'', cons(y'', l''))
REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, cons(rev1(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))
REV2(x, cons(y, l)) -> REV2(y, l)

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

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

REV2(x, cons(y, l)) -> REV2(y, l)
two new Dependency Pairs are created:

REV2(x, cons(y0, cons(y'', l''))) -> REV2(y0, cons(y'', l''))
REV2(x, cons(y', cons(y0'', cons(y'''', l'''')))) -> REV2(y', cons(y0'', cons(y'''', l'''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 9`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

REV2(x, cons(y', cons(y0'', cons(y'''', l'''')))) -> REV2(y', cons(y0'', cons(y'''', l'''')))
REV2(x, cons(y0, cons(y'', l''))) -> REV2(y0, cons(y'', l''))
REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, cons(rev1(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))
REV(cons(x'', cons(y'', l''))) -> REV2(x'', cons(y'', l''))

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

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

REV(cons(x'', cons(y'', l''))) -> REV2(x'', cons(y'', l''))
two new Dependency Pairs are created:

REV(cons(x''', cons(y''0, cons(y'''', l'''')))) -> REV2(x''', cons(y''0, cons(y'''', l'''')))
REV(cons(x''', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))) -> REV2(x''', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 10`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

REV(cons(x''', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))) -> REV2(x''', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))
REV2(x, cons(y0, cons(y'', l''))) -> REV2(y0, cons(y'', l''))
REV(cons(x''', cons(y''0, cons(y'''', l'''')))) -> REV2(x''', cons(y''0, cons(y'''', l'''')))
REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, cons(rev1(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))
REV2(x, cons(y', cons(y0'', cons(y'''', l'''')))) -> REV2(y', cons(y0'', cons(y'''', l'''')))

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

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

REV2(x, cons(y0, cons(y'', l''))) -> REV2(y0, cons(y'', l''))
two new Dependency Pairs are created:

REV2(x, cons(y0'', cons(y''0, cons(y'''', l'''')))) -> REV2(y0'', cons(y''0, cons(y'''', l'''')))
REV2(x, cons(y0', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))) -> REV2(y0', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 11`
`                 ↳Polynomial Ordering`

Dependency Pairs:

REV2(x, cons(y0', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))) -> REV2(y0', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))
REV2(x, cons(y0'', cons(y''0, cons(y'''', l'''')))) -> REV2(y0'', cons(y''0, cons(y'''', l'''')))
REV2(x, cons(y', cons(y0'', cons(y'''', l'''')))) -> REV2(y', cons(y0'', cons(y'''', l'''')))
REV(cons(x''', cons(y''0, cons(y'''', l'''')))) -> REV2(x''', cons(y''0, cons(y'''', l'''')))
REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, cons(rev1(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))
REV(cons(x''', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))) -> REV2(x''', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

REV2(x, cons(y0', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))) -> REV2(y0', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))
REV2(x, cons(y0'', cons(y''0, cons(y'''', l'''')))) -> REV2(y0'', cons(y''0, cons(y'''', l'''')))
REV2(x, cons(y', cons(y0'', cons(y'''', l'''')))) -> REV2(y', cons(y0'', cons(y'''', l'''')))
REV(cons(x''', cons(y''0, cons(y'''', l'''')))) -> REV2(x''', cons(y''0, cons(y'''', l'''')))
REV(cons(x''', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))) -> REV2(x''', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))
rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(rev2(x1, x2)) =  x2 POL(rev(x1)) =  x1 POL(0) =  0 POL(REV(x1)) =  x1 POL(cons(x1, x2)) =  1 + x2 POL(rev1(x1, x2)) =  0 POL(REV2(x1, x2)) =  x2 POL(nil) =  0 POL(s(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 12`
`                 ↳Dependency Graph`

Dependency Pair:

REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, cons(rev1(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))

Rules:

rev(nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, nil) -> nil
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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