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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs.


   R
DPs
       →DP Problem 1
Narrowing Transformation
       →DP Problem 2
Remaining


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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

F(a, f(b, x)) -> F(a, f(a, f(a, x)))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Narrowing Transformation
       →DP Problem 2
Remaining


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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

F(a, f(b, x)) -> F(a, f(a, x))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Nar
             ...
               →DP Problem 4
Narrowing Transformation
       →DP Problem 2
Remaining


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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

F(a, f(b, f(b, x''))) -> F(a, f(a, f(a, f(a, f(a, x'')))))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Nar
             ...
               →DP Problem 5
Narrowing Transformation
       →DP Problem 2
Remaining


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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

F(a, f(b, f(b, x''))) -> F(a, f(a, f(a, f(a, x''))))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Nar
             ...
               →DP Problem 6
Forward Instantiation Transformation
       →DP Problem 2
Remaining


Dependency Pairs:

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


Rules:


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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Nar
             ...
               →DP Problem 7
Polynomial Ordering
       →DP Problem 2
Remaining


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

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


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

resulting in one new DP problem.



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




The following remains to be proven:


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




The following remains to be proven:

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