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 rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(unit(x1)) =  x1 POL(rev(x1)) =  x1 POL(++'(x1, x2)) =  1 + x1 POL(flatten(x1)) =  x1 POL(++(x1, x2)) =  1 + x1 + x2 POL(nil) =  0

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

Additionally, the following rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(unit(x1)) =  1 + x1 POL(rev(x1)) =  x1 POL(flatten(x1)) =  0 POL(++(x1, x2)) =  x1 + x2 POL(FLATTEN(x1)) =  1 + x1 POL(nil) =  0

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 Pairs:

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

Additionally, the following rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(unit(x1)) =  x1 POL(rev(x1)) =  x1 POL(flatten(x1)) =  x1 POL(++(x1, x2)) =  1 + x1 + x2 POL(FLATTEN(x1)) =  1 + x1 POL(nil) =  0

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)

Additionally, the following rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(unit(x1)) =  x1 POL(rev(x1)) =  x1 POL(REV(x1)) =  1 + x1 POL(flatten(x1)) =  x1 POL(++(x1, x2)) =  1 + x1 + x2 POL(nil) =  0

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