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


The following usable rules for innermost 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))=  x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
++'(x1, x2) -> ++'(x1, x2)
++(x1, x2) -> ++(x1, x2)
.(x1, x2) -> .(x1, x2)


   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
                 ↳Argument Filtering and 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 that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(++'(x1, x2))=  1 + x1 + x2  
  POL(.(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
++'(x1, x2) -> ++'(x1, x2)
.(x1, x2) -> .(x1, x2)


   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