Term Rewriting System R:
[x]
f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(s(s(x))) -> F(f(s(x)))
F(s(s(x))) -> F(s(x))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

F(s(s(x))) -> F(s(x))
F(s(s(x))) -> F(f(s(x)))


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Strategy:

innermost




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

F(s(s(x))) -> F(f(s(x)))
two new Dependency Pairs are created:

F(s(s(0))) -> F(s(0))
F(s(s(s(x'')))) -> F(f(f(s(x''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Narrowing Transformation


Dependency Pairs:

F(s(s(s(x'')))) -> F(f(f(s(x''))))
F(s(s(x))) -> F(s(x))


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Strategy:

innermost




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

F(s(s(s(x'')))) -> F(f(f(s(x''))))
two new Dependency Pairs are created:

F(s(s(s(0)))) -> F(f(s(0)))
F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 3
                 ↳Rewriting Transformation


Dependency Pairs:

F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))
F(s(s(s(0)))) -> F(f(s(0)))
F(s(s(x))) -> F(s(x))


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Strategy:

innermost




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

F(s(s(s(0)))) -> F(f(s(0)))
one new Dependency Pair is created:

F(s(s(s(0)))) -> F(s(0))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

F(s(s(x))) -> F(s(x))
F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Strategy:

innermost




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

F(s(s(x))) -> F(s(x))
two new Dependency Pairs are created:

F(s(s(s(x'')))) -> F(s(s(x'')))
F(s(s(s(s(s(x''')))))) -> F(s(s(s(s(x''')))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 5
                 ↳Forward Instantiation Transformation


Dependency Pairs:

F(s(s(s(s(s(x''')))))) -> F(s(s(s(s(x''')))))
F(s(s(s(x'')))) -> F(s(s(x'')))
F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Strategy:

innermost




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

F(s(s(s(x'')))) -> F(s(s(x'')))
three new Dependency Pairs are created:

F(s(s(s(s(s(x'''')))))) -> F(s(s(s(s(x'''')))))
F(s(s(s(s(x''''))))) -> F(s(s(s(x''''))))
F(s(s(s(s(s(s(x'''''))))))) -> F(s(s(s(s(s(x'''''))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 6
                 ↳Argument Filtering and Ordering


Dependency Pairs:

F(s(s(s(s(s(s(x'''''))))))) -> F(s(s(s(s(s(x'''''))))))
F(s(s(s(s(x''''))))) -> F(s(s(s(x''''))))
F(s(s(s(s(s(x'''')))))) -> F(s(s(s(s(x'''')))))
F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))
F(s(s(s(s(s(x''')))))) -> F(s(s(s(s(x''')))))


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

F(s(s(s(s(s(s(x'''''))))))) -> F(s(s(s(s(s(x'''''))))))
F(s(s(s(s(x''''))))) -> F(s(s(s(x''''))))
F(s(s(s(s(s(x'''')))))) -> F(s(s(s(s(x'''')))))
F(s(s(s(s(s(x''')))))) -> F(s(s(s(s(x''')))))


The following usable rules for innermost can be oriented:

f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(s(x1))=  1 + x1  
  POL(F(x1))=  1 + x1  
  POL(f(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
s(x1) -> s(x1)
f(x1) -> f(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 7
                 ↳Argument Filtering and Ordering


Dependency Pair:

F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))


The following usable rules for innermost can be oriented:

f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(s(x1))=  1 + x1  
  POL(F(x1))=  1 + x1  
  POL(f)=  1  

resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
s(x1) -> s(x1)
f(x1) -> f


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


Dependency Pair:


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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