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))

Innermost 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`
`         ↳Forward Instantiation Transformation`

Dependency Pairs:

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

Strategy:

innermost

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

++'(.(x, y), z) -> ++'(y, z)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

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

Strategy:

innermost

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

++'(++(x, y), z) -> ++'(y, z)
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

++'(.(x, .(x'', y'')), z'') -> ++'(.(x'', y''), z'')
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 4`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

++'(++(x, .(x'', .(x'''', y''''))), z') -> ++'(.(x'', .(x'''', y'''')), z')
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Polynomial Ordering`

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS 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)) =  1 + x1 + x2 POL(++(x1, x2)) =  1 + x1 + x2 POL(nil) =  1 POL(.(x1, x2)) =  x2

resulting in one new DP problem.

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

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Polynomial Ordering`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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(++'(x1, x2)) =  1 + x1 POL(.(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 8`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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