Term Rewriting System R:
[x]
f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(a, x) -> F(b, f(c, x))
F(a, x) -> F(c, x)
F(a, f(b, x)) -> F(b, f(a, x))
F(a, f(b, x)) -> F(a, x)
F(d, f(c, x)) -> F(d, f(a, x))
F(d, f(c, x)) -> F(a, x)
F(a, f(c, x)) -> F(c, f(a, x))
F(a, f(c, x)) -> F(a, x)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

F(a, f(c, x)) -> F(a, x)
F(a, f(b, x)) -> F(a, x)


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

F(a, f(c, x)) -> F(a, x)
F(a, f(b, x)) -> F(a, x)


There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c)=  0  
  POL(b)=  0  
  POL(a)=  0  
  POL(F(x1, x2))=  x1 + x2  
  POL(f(x1, x2))=  1 + x1 + x2  

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Nar


Dependency Pair:


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

F(d, f(c, x)) -> F(d, f(a, x))


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))


Strategy:

innermost




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

F(d, f(c, x)) -> F(d, f(a, x))
three new Dependency Pairs are created:

F(d, f(c, x'')) -> F(d, f(b, f(c, x'')))
F(d, f(c, f(b, x''))) -> F(d, f(b, f(a, x'')))
F(d, f(c, f(c, x''))) -> F(d, f(c, f(a, x'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Narrowing Transformation


Dependency Pairs:

F(d, f(c, f(c, x''))) -> F(d, f(c, f(a, x'')))
F(d, f(c, f(b, x''))) -> F(d, f(b, f(a, x'')))
F(d, f(c, x'')) -> F(d, f(b, f(c, x'')))


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))


Strategy:

innermost




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

F(d, f(c, x'')) -> F(d, f(b, f(c, x'')))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 5
                 ↳Argument Filtering and Ordering


Dependency Pairs:

F(d, f(c, f(b, x''))) -> F(d, f(b, f(a, x'')))
F(d, f(c, f(c, x''))) -> F(d, f(c, f(a, x'')))


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(d, f(c, f(b, x''))) -> F(d, f(b, f(a, x'')))


The following usable rules for innermost can be oriented:

f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))


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

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 6
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

F(d, f(c, f(c, x''))) -> F(d, f(c, f(a, x'')))


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:00 minutes