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)))

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

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)))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

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)))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(rev2(x1, x2)) =  x1 + x2 POL(rev(x1)) =  x1 POL(0) =  0 POL(REV1(x1, x2)) =  1 + x1 + x2 POL(cons(x1, x2)) =  1 + x1 + x2 POL(rev1) =  0 POL(nil) =  0 POL(s) =  0

resulting in one new DP problem.
Used Argument Filtering System:
REV1(x1, x2) -> REV1(x1, x2)
cons(x1, x2) -> cons(x1, x2)
rev(x1) -> rev(x1)
rev1(x1, x2) -> rev1
rev2(x1, x2) -> rev2(x1, x2)
s(x1) -> s

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

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)))

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, 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)))

The following dependency pairs can be strictly oriented:

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

The following rules 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))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)

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

resulting in one new DP problem.
Used Argument Filtering System:
REV2(x1, x2) -> REV2(x1, x2)
REV(x1) -> REV(x1)
cons(x1, x2) -> cons(x1, x2)
rev2(x1, x2) -> rev2(x1, x2)
rev(x1) -> rev(x1)
rev1(x1, x2) -> rev1
s(x1) -> s

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

Dependency Pair:

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)))

Using the Dependency Graph resulted in no new DP problems.

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