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
                 ↳Forward Instantiation Transformation


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




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''''0, .(x'''''0, .(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, .(x'''''''', y''''''))), z'''''')
++'(.(x, .(x''''0, .(x'''''0, ++(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, ++(x'''''''', y''''''))), z'''''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 6
                 ↳Forward Instantiation Transformation


Dependency Pairs:

++'(.(x, .(x''''0, .(x'''''0, ++(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, ++(x'''''''', y''''''))), z'''''')
++'(.(x, .(x''''0, .(x'''''0, .(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, .(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'''')
++'(++(x, .(x''', .(x''''', ++(x'''''''', y'''''')))), z'') -> ++'(.(x''', .(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''''', .(x'''''''', y'''''')))), z'') -> ++'(.(x''', .(x''''', .(x'''''''', y''''''))), z'')
two new Dependency Pairs are created:

++'(++(x, .(x'''', .(x'''''', .(x'''''''''', .(x''''''''''', y''''''''))))), z''') -> ++'(.(x'''', .(x'''''', .(x'''''''''', .(x''''''''''', y'''''''')))), z''')
++'(++(x, .(x'''', .(x'''''', .(x'''''''''', ++(x''''''''''', y''''''''))))), z''') -> ++'(.(x'''', .(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 7
                 ↳Forward Instantiation Transformation


Dependency Pairs:

++'(++(x, .(x'''', .(x'''''', .(x'''''''''', ++(x''''''''''', y''''''''))))), z''') -> ++'(.(x'''', .(x'''''', .(x'''''''''', ++(x''''''''''', y'''''''')))), z''')
++'(.(x, .(x''''0, .(x'''''0, .(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, .(x'''''''', y''''''))), z'''''')
++'(++(x, .(x'''', .(x'''''', .(x'''''''''', .(x''''''''''', y''''''''))))), z''') -> ++'(.(x'''', .(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'')
++'(.(x, ++(x'', y'')), z'') -> ++'(++(x'', y''), z'')
++'(.(x, .(x'''', ++(x''''', y''''))), z'''') -> ++'(.(x'''', ++(x''''', y'''')), z'''')
++'(.(x, .(x''''0, .(x'''''0, ++(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, ++(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''''0, .(x'''''0, .(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, .(x'''''''', y''''''))), z'''''')
two new Dependency Pairs are created:

++'(.(x, .(x''''0'', .(x'''''0'', .(x'''''''''', .(x''''''''''', y''''''''))))), z'''''''') -> ++'(.(x''''0'', .(x'''''0'', .(x'''''''''', .(x''''''''''', y'''''''')))), z'''''''')
++'(.(x, .(x''''0'', .(x'''''0'', .(x'''''''''', ++(x''''''''''', y''''''''))))), z'''''''') -> ++'(.(x''''0'', .(x'''''0'', .(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 8
                 ↳Forward Instantiation Transformation


Dependency Pairs:

++'(.(x, .(x''''0'', .(x'''''0'', .(x'''''''''', ++(x''''''''''', y''''''''))))), z'''''''') -> ++'(.(x''''0'', .(x'''''0'', .(x'''''''''', ++(x''''''''''', y'''''''')))), z'''''''')
++'(.(x, .(x''''0'', .(x'''''0'', .(x'''''''''', .(x''''''''''', y''''''''))))), z'''''''') -> ++'(.(x''''0'', .(x'''''0'', .(x'''''''''', .(x''''''''''', y'''''''')))), z'''''''')
++'(++(x, .(x'''', .(x'''''', .(x'''''''''', .(x''''''''''', y''''''''))))), z''') -> ++'(.(x'''', .(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'')
++'(.(x, ++(x'', y'')), z'') -> ++'(++(x'', y''), z'')
++'(.(x, .(x'''', ++(x''''', y''''))), z'''') -> ++'(.(x'''', ++(x''''', y'''')), z'''')
++'(.(x, .(x''''0, .(x'''''0, ++(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, ++(x'''''''', y''''''))), z'''''')
++'(++(x, .(x'''', .(x'''''', .(x'''''''''', ++(x''''''''''', y''''''''))))), z''') -> ++'(.(x'''', .(x'''''', .(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'''''', .(x'''''''''', .(x''''''''''', y''''''''))))), z''') -> ++'(.(x'''', .(x'''''', .(x'''''''''', .(x''''''''''', y'''''''')))), z''')
two new Dependency Pairs are created:

++'(++(x, .(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y'''''''''')))))), z'''') -> ++'(.(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y''''''''''))))), z'''')
++'(++(x, .(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y'''''''''')))))), z'''') -> ++'(.(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y''''''''''))))), z'''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 9
                 ↳Forward Instantiation Transformation


Dependency Pairs:

++'(++(x, .(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y'''''''''')))))), z'''') -> ++'(.(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y''''''''''))))), z'''')
++'(.(x, .(x''''0'', .(x'''''0'', .(x'''''''''', .(x''''''''''', y''''''''))))), z'''''''') -> ++'(.(x''''0'', .(x'''''0'', .(x'''''''''', .(x''''''''''', y'''''''')))), z'''''''')
++'(++(x, .(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y'''''''''')))))), z'''') -> ++'(.(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y''''''''''))))), z'''')
++'(++(x, .(x'''', .(x'''''', .(x'''''''''', ++(x''''''''''', y''''''''))))), z''') -> ++'(.(x'''', .(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'')
++'(.(x, ++(x'', y'')), z'') -> ++'(++(x'', y''), z'')
++'(.(x, .(x'''', ++(x''''', y''''))), z'''') -> ++'(.(x'''', ++(x''''', y'''')), z'''')
++'(.(x, .(x''''0, .(x'''''0, ++(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, ++(x'''''''', y''''''))), z'''''')
++'(.(x, .(x''''0'', .(x'''''0'', .(x'''''''''', ++(x''''''''''', y''''''''))))), z'''''''') -> ++'(.(x''''0'', .(x'''''0'', .(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''''0'', .(x'''''0'', .(x'''''''''', .(x''''''''''', y''''''''))))), z'''''''') -> ++'(.(x''''0'', .(x'''''0'', .(x'''''''''', .(x''''''''''', y'''''''')))), z'''''''')
two new Dependency Pairs are created:

++'(.(x, .(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y'''''''''')))))), z'''''''''') -> ++'(.(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y''''''''''))))), z'''''''''')
++'(.(x, .(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y'''''''''')))))), z'''''''''') -> ++'(.(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y''''''''''))))), z'''''''''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 10
                 ↳Forward Instantiation Transformation


Dependency Pairs:

++'(.(x, .(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y'''''''''')))))), z'''''''''') -> ++'(.(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y''''''''''))))), z'''''''''')
++'(.(x, .(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y'''''''''')))))), z'''''''''') -> ++'(.(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y''''''''''))))), z'''''''''')
++'(++(x, .(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y'''''''''')))))), z'''') -> ++'(.(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y''''''''''))))), z'''')
++'(++(x, .(x'''', .(x'''''', .(x'''''''''', ++(x''''''''''', y''''''''))))), z''') -> ++'(.(x'''', .(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'')
++'(.(x, ++(x'', y'')), z'') -> ++'(++(x'', y''), z'')
++'(.(x, .(x'''', ++(x''''', y''''))), z'''') -> ++'(.(x'''', ++(x''''', y'''')), z'''')
++'(.(x, .(x''''0, .(x'''''0, ++(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, ++(x'''''''', y''''''))), z'''''')
++'(.(x, .(x''''0'', .(x'''''0'', .(x'''''''''', ++(x''''''''''', y''''''''))))), z'''''''') -> ++'(.(x''''0'', .(x'''''0'', .(x'''''''''', ++(x''''''''''', y'''''''')))), z'''''''')
++'(++(x, .(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y'''''''''')))))), z'''') -> ++'(.(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, ++(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''''''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y'''''''''')))))), z'''') -> ++'(.(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y''''''''''))))), z'''')
two new Dependency Pairs are created:

++'(++(x, .(x'''''', .(x'''''''', .(x''''''''''0'', .(x'''''''''''0'', .(x''''''''''''''0, .(x'''''''''''''''', y''''''''''''))))))), z''''') -> ++'(.(x'''''', .(x'''''''', .(x''''''''''0'', .(x'''''''''''0'', .(x''''''''''''''0, .(x'''''''''''''''', y'''''''''''')))))), z''''')
++'(++(x, .(x'''''', .(x'''''''', .(x''''''''''0'', .(x'''''''''''0'', .(x''''''''''''''0, ++(x'''''''''''''''', y''''''''''''))))))), z''''') -> ++'(.(x'''''', .(x'''''''', .(x''''''''''0'', .(x'''''''''''0'', .(x''''''''''''''0, ++(x'''''''''''''''', y'''''''''''')))))), z''''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 11
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

++'(++(x, .(x'''''', .(x'''''''', .(x''''''''''0'', .(x'''''''''''0'', .(x''''''''''''''0, ++(x'''''''''''''''', y''''''''''''))))))), z''''') -> ++'(.(x'''''', .(x'''''''', .(x''''''''''0'', .(x'''''''''''0'', .(x''''''''''''''0, ++(x'''''''''''''''', y'''''''''''')))))), z''''')
++'(.(x, .(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y'''''''''')))))), z'''''''''') -> ++'(.(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, .(x'''''''''''''', y''''''''''))))), z'''''''''')
++'(++(x, .(x'''''', .(x'''''''', .(x''''''''''0'', .(x'''''''''''0'', .(x''''''''''''''0, .(x'''''''''''''''', y''''''''''''))))))), z''''') -> ++'(.(x'''''', .(x'''''''', .(x''''''''''0'', .(x'''''''''''0'', .(x''''''''''''''0, .(x'''''''''''''''', y'''''''''''')))))), z''''')
++'(++(x, .(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y'''''''''')))))), z'''') -> ++'(.(x''''', .(x''''''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y''''''''''))))), z'''')
++'(++(x, .(x'''', .(x'''''', .(x'''''''''', ++(x''''''''''', y''''''''))))), z''') -> ++'(.(x'''', .(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'')
++'(.(x, ++(x'', y'')), z'') -> ++'(++(x'', y''), z'')
++'(.(x, .(x'''', ++(x''''', y''''))), z'''') -> ++'(.(x'''', ++(x''''', y'''')), z'''')
++'(.(x, .(x''''0, .(x'''''0, ++(x'''''''', y'''''')))), z'''''') -> ++'(.(x''''0, .(x'''''0, ++(x'''''''', y''''''))), z'''''')
++'(.(x, .(x''''0'', .(x'''''0'', .(x'''''''''', ++(x''''''''''', y''''''''))))), z'''''''') -> ++'(.(x''''0'', .(x'''''0'', .(x'''''''''', ++(x''''''''''', y'''''''')))), z'''''''')
++'(.(x, .(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y'''''''''')))))), z'''''''''') -> ++'(.(x''''0'''', .(x'''''0'''', .(x''''''''''0, .(x'''''''''''0, ++(x'''''''''''''', y''''''''''))))), z'''''''''')


Rules:


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


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:01 minutes