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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`

Dependency Pairs:

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

Rules:

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

++(nil, y) -> y
++(x, nil) -> x
++(.(x, y), z) -> .(x, ++(y, z))
++(++(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 POL(.(x1, x2)) =  x2

resulting in one new DP problem.

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

Dependency Pair:

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

Rules:

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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

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

resulting in one new DP problem.

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

Dependency Pair:

Rules:

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