Term Rewriting System R:
[x, y, z]
flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

FLATTEN(unit(x)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> ++'(flatten(x), flatten(y))
FLATTEN(++(x, y)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> FLATTEN(y)
FLATTEN(++(unit(x), y)) -> ++'(flatten(x), flatten(y))
FLATTEN(++(unit(x), y)) -> FLATTEN(x)
FLATTEN(++(unit(x), y)) -> FLATTEN(y)
REV(++(x, y)) -> ++'(rev(y), rev(x))
REV(++(x, y)) -> REV(y)
REV(++(x, y)) -> REV(x)
++'(++(x, y), z) -> ++'(x, ++(y, z))
++'(++(x, y), z) -> ++'(y, z)

Furthermore, R contains three SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pairs:

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

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

The following dependency pairs can be strictly oriented:

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

Additionally, the following usable rules using the Ce-refinement can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(++'(x1, x2)) =  x1 POL(++(x1, x2)) =  1 + x1 + x2 POL(nil) =  1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pairs:

FLATTEN(++(unit(x), y)) -> FLATTEN(y)
FLATTEN(++(unit(x), y)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> FLATTEN(y)
FLATTEN(++(x, y)) -> FLATTEN(x)
FLATTEN(unit(x)) -> FLATTEN(x)

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

The following dependency pairs can be strictly oriented:

FLATTEN(++(unit(x), y)) -> FLATTEN(y)
FLATTEN(++(unit(x), y)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> FLATTEN(y)
FLATTEN(++(x, y)) -> FLATTEN(x)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(unit(x1)) =  x1 POL(FLATTEN(x1)) =  1 + x1 POL(++(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

FLATTEN(unit(x)) -> FLATTEN(x)

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

The following dependency pair can be strictly oriented:

FLATTEN(unit(x)) -> FLATTEN(x)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(unit(x1)) =  1 + x1 POL(FLATTEN(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Polo`
`             ...`
`               →DP Problem 6`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polynomial Ordering`

Dependency Pairs:

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

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

The following dependency pairs can be strictly oriented:

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

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(REV(x1)) =  x1 POL(++(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 7`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Using the Dependency Graph resulted in no new DP problems.

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