Term Rewriting System R:
[x, y, z]
f(+(x, 0)) -> f(x)
+(x, +(y, z)) -> +(+(x, y), z)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


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


Dependency Pair:

F(+(x, 0)) -> F(x)


Rules:


f(+(x, 0)) -> f(x)
+(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

F(+(x, 0)) -> F(x)
one new Dependency Pair is created:

F(+(+(x'', 0), 0)) -> F(+(x'', 0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

F(+(+(x'', 0), 0)) -> F(+(x'', 0))


Rules:


f(+(x, 0)) -> f(x)
+(x, +(y, z)) -> +(+(x, y), z)


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(+(+(x'', 0), 0)) -> F(+(x'', 0))


The following usable rule for innermost can be oriented:

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


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

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


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳AFS
             ...
               →DP Problem 4
                 ↳Dependency Graph
       →DP Problem 2
         ↳FwdInst


Dependency Pair:


Rules:


f(+(x, 0)) -> f(x)
+(x, +(y, z)) -> +(+(x, y), z)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

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


Rules:


f(+(x, 0)) -> f(x)
+(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)) -> +'(x, y)
one new Dependency Pair is created:

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

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


Rules:


f(+(x, 0)) -> f(x)
+(x, +(y, z)) -> +(+(x, y), z)


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


The following usable rule for innermost can be oriented:

+(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  

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 5
             ↳AFS
             ...
               →DP Problem 6
                 ↳Dependency Graph


Dependency Pair:


Rules:


f(+(x, 0)) -> f(x)
+(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