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
         ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining


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




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

F(a, f(b, x)) -> F(a, x)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

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




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

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




The following dependency pairs can be strictly oriented:

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


Additionally, 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)=  1  
  POL(a)=  1  
  POL(f(x1, x2))=  x1 + x2  
  POL(F(x1, x2))=  1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

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




Using the Dependency Graph the DP problem was split into 1 DP problems.


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




The following remains to be proven:


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




The following remains to be proven:

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