Termination Proof using AProVE

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.

     ↳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.

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 5

                 ↳Argument Filtering and 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''''', 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'''')

The following usable rules for innermost w.r.t. to the AFS can be oriented:

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

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

{., ++}

resulting in one new DP problem.
Used Argument Filtering System:

++'(x₁, x₂) -> ++'(x₁, x₂)
++(x₁, x₂) -> ++(x₁, x₂)
.(x₁, x₂) -> .(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 6

                 ↳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