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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.


   R
DPs
       →DP Problem 1
Narrowing Transformation


Dependency Pairs:

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


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



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


Dependency Pairs:

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


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



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


Dependency Pairs:

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


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



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


Dependency Pairs:

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


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



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


Dependency Pairs:

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


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(y', z')) -> F(b, f(a, f(c, f(f(b, y'), z'))))
three new Dependency Pairs are created:

F(f(f(a, b), c), f(y'', z'')) -> F(b, f(f(a, c), f(f(b, y''), z'')))
F(f(f(a, b), c), f(y'', z'')) -> F(b, f(a, f(f(c, f(b, y'')), z'')))
F(f(f(a, b), c), f(f(y'', z''), z')) -> F(b, f(a, f(c, f(f(f(b, y''), z''), z'))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 6
Narrowing Transformation


Dependency Pairs:

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


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(y', z')) -> F(a, f(c, f(f(b, y'), z')))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(y'', z'')) -> F(a, f(f(c, f(b, y'')), z''))
F(f(f(a, b), c), f(f(y'', z''), z')) -> F(a, f(c, f(f(f(b, y''), z''), z')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 7
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


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

F(f(f(a, b), c), f(y'', z'')) -> F(b, f(y'', z''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 8
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(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, f(y, z)) -> F(x, y)
five new Dependency Pairs are created:

F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 9
Polynomial Ordering


Dependency Pairs:

F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(f(f(a, b), c), f(f(y'', z''), z')) -> F(a, f(c, f(f(f(b, y''), z''), z')))
F(f(f(a, b), c), f(y'', z'')) -> F(a, f(f(c, f(b, y'')), z''))
F(f(f(a, b), c), f(f(y'', z''), z')) -> F(b, f(a, f(c, f(f(f(b, y''), z''), z'))))
F(f(f(a, b), c), f(y'', z'')) -> F(b, f(a, f(f(c, f(b, y'')), z'')))
F(f(f(a, b), c), f(y'', z'')) -> F(b, f(f(a, c), f(f(b, y''), z'')))
F(f(f(a, b), c), f(y', z')) -> F(b, f(a, f(f(f(c, b), y'), z')))
F(f(f(a, b), c), x''') -> F(b, f(f(a, f(c, b)), x'''))
F(f(f(a, b), c), f(y', z')) -> F(b, f(f(a, c), f(f(b, y'), z')))
F(f(f(a, b), c), x''') -> F(b, f(f(f(a, c), b), x'''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x) -> F(c, f(b, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(y'', z'')) -> F(b, f(y'', z''))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules for innermost can be oriented:

f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 10
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(f(f(a, b), c), f(f(y'', z''), z')) -> F(a, f(c, f(f(f(b, y''), z''), z')))
F(f(f(a, b), c), f(y'', z'')) -> F(a, f(f(c, f(b, y'')), z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(x, f(y, z)) -> F(f(x, y), z)


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(y'', z'')) -> F(a, f(f(c, f(b, y'')), z''))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(y', z'), z'')) -> F(a, f(f(c, f(f(b, y'), z')), z''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 11
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(y', z'), z'')) -> F(a, f(f(c, f(f(b, y'), z')), z''))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(f(f(a, b), c), f(f(y'', z''), z')) -> F(a, f(c, f(f(f(b, y''), z''), z')))
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(f(y'', z''), z')) -> F(a, f(c, f(f(f(b, y''), z''), z')))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(f(y''', z'''), z'0)) -> F(a, f(f(c, f(f(b, y'''), z''')), z'0))
F(f(f(a, b), c), f(f(f(y', z0), z''), z')) -> F(a, f(c, f(f(f(f(b, y'), z0), z''), z')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 12
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(f(y', z0), z''), z')) -> F(a, f(c, f(f(f(f(b, y'), z0), z''), z')))
F(f(f(a, b), c), f(f(y''', z'''), z'0)) -> F(a, f(f(c, f(f(b, y'''), z''')), z'0))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(y', z'), z'')) -> F(a, f(f(c, f(f(b, y'), z')), z''))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(f(y', z'), z'')) -> F(a, f(f(c, f(f(b, y'), z')), z''))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(f(y'', z'''), z'')) -> F(a, f(f(f(c, f(b, y'')), z'''), z''))
F(f(f(a, b), c), f(f(f(y'', z0), z'), z'')) -> F(a, f(f(c, f(f(f(b, y''), z0), z')), z''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 13
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(f(y'', z0), z'), z'')) -> F(a, f(f(c, f(f(f(b, y''), z0), z')), z''))
F(f(f(a, b), c), f(f(y'', z'''), z'')) -> F(a, f(f(f(c, f(b, y'')), z'''), z''))
F(f(f(a, b), c), f(f(y''', z'''), z'0)) -> F(a, f(f(c, f(f(b, y'''), z''')), z'0))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(f(y', z0), z''), z')) -> F(a, f(c, f(f(f(f(b, y'), z0), z''), z')))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(f(y''', z'''), z'0)) -> F(a, f(f(c, f(f(b, y'''), z''')), z'0))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(f(y'''', z''''), z'0)) -> F(a, f(f(f(c, f(b, y'''')), z''''), z'0))
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y'), z'), z''')), z'0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 14
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(f(y', z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y'), z'), z''')), z'0))
F(f(f(a, b), c), f(f(y'''', z''''), z'0)) -> F(a, f(f(f(c, f(b, y'''')), z''''), z'0))
F(f(f(a, b), c), f(f(y'', z'''), z'')) -> F(a, f(f(f(c, f(b, y'')), z'''), z''))
F(f(f(a, b), c), f(f(f(y', z0), z''), z')) -> F(a, f(c, f(f(f(f(b, y'), z0), z''), z')))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(f(y'', z0), z'), z'')) -> F(a, f(f(c, f(f(f(b, y''), z0), z')), z''))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(f(f(y', z0), z''), z')) -> F(a, f(c, f(f(f(f(b, y'), z0), z''), z')))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(f(f(y'', z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y''), z0'), z''')), z'0))
F(f(f(a, b), c), f(f(f(f(y'', z1), z0), z''), z')) -> F(a, f(c, f(f(f(f(f(b, y''), z1), z0), z''), z')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 15
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(f(f(y'', z1), z0), z''), z')) -> F(a, f(c, f(f(f(f(f(b, y''), z1), z0), z''), z')))
F(f(f(a, b), c), f(f(f(y'', z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y''), z0'), z''')), z'0))
F(f(f(a, b), c), f(f(y'''', z''''), z'0)) -> F(a, f(f(f(c, f(b, y'''')), z''''), z'0))
F(f(f(a, b), c), f(f(f(y'', z0), z'), z'')) -> F(a, f(f(c, f(f(f(b, y''), z0), z')), z''))
F(f(f(a, b), c), f(f(y'', z'''), z'')) -> F(a, f(f(f(c, f(b, y'')), z'''), z''))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y'), z'), z''')), z'0))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(f(y'', z'''), z'')) -> F(a, f(f(f(c, f(b, y'')), z'''), z''))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(f(y''', z'''), z'')) -> F(a, f(f(f(f(c, b), y'''), z'''), z''))
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'), z')), z'''), z''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 16
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(f(y', z'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'), z')), z'''), z''))
F(f(f(a, b), c), f(f(y''', z'''), z'')) -> F(a, f(f(f(f(c, b), y'''), z'''), z''))
F(f(f(a, b), c), f(f(f(y'', z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y''), z0'), z''')), z'0))
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y'), z'), z''')), z'0))
F(f(f(a, b), c), f(f(y'''', z''''), z'0)) -> F(a, f(f(f(c, f(b, y'''')), z''''), z'0))
F(f(f(a, b), c), f(f(f(y'', z0), z'), z'')) -> F(a, f(f(c, f(f(f(b, y''), z0), z')), z''))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(f(f(y'', z1), z0), z''), z')) -> F(a, f(c, f(f(f(f(f(b, y''), z1), z0), z''), z')))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(f(f(y'', z0), z'), z'')) -> F(a, f(f(c, f(f(f(b, y''), z0), z')), z''))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(f(f(y''', z0'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'''), z0')), z'''), z''))
F(f(f(a, b), c), f(f(f(f(y', z1), z0), z'), z'')) -> F(a, f(f(c, f(f(f(f(b, y'), z1), z0), z')), z''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 17
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(f(f(y', z1), z0), z'), z'')) -> F(a, f(f(c, f(f(f(f(b, y'), z1), z0), z')), z''))
F(f(f(a, b), c), f(f(f(y''', z0'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'''), z0')), z'''), z''))
F(f(f(a, b), c), f(f(y''', z'''), z'')) -> F(a, f(f(f(f(c, b), y'''), z'''), z''))
F(f(f(a, b), c), f(f(f(f(y'', z1), z0), z''), z')) -> F(a, f(c, f(f(f(f(f(b, y''), z1), z0), z''), z')))
F(f(f(a, b), c), f(f(f(y'', z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y''), z0'), z''')), z'0))
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y'), z'), z''')), z'0))
F(f(f(a, b), c), f(f(y'''', z''''), z'0)) -> F(a, f(f(f(c, f(b, y'''')), z''''), z'0))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'), z')), z'''), z''))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(f(y'''', z''''), z'0)) -> F(a, f(f(f(c, f(b, y'''')), z''''), z'0))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(f(y''''', z''''), z'0)) -> F(a, f(f(f(f(c, b), y'''''), z''''), z'0))
F(f(f(a, b), c), f(f(f(y', z'), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y'), z')), z''''), z'0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 18
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(f(y', z'), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y'), z')), z''''), z'0))
F(f(f(a, b), c), f(f(y''''', z''''), z'0)) -> F(a, f(f(f(f(c, b), y'''''), z''''), z'0))
F(f(f(a, b), c), f(f(f(y''', z0'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'''), z0')), z'''), z''))
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'), z')), z'''), z''))
F(f(f(a, b), c), f(f(y''', z'''), z'')) -> F(a, f(f(f(f(c, b), y'''), z'''), z''))
F(f(f(a, b), c), f(f(f(f(y'', z1), z0), z''), z')) -> F(a, f(c, f(f(f(f(f(b, y''), z1), z0), z''), z')))
F(f(f(a, b), c), f(f(f(y'', z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y''), z0'), z''')), z'0))
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y'), z'), z''')), z'0))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(f(f(y', z1), z0), z'), z'')) -> F(a, f(f(c, f(f(f(f(b, y'), z1), z0), z')), z''))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(f(f(y', z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y'), z'), z''')), z'0))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(f(f(y'', z''), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y''), z'')), z''''), z'0))
F(f(f(a, b), c), f(f(f(f(y'', z''), z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(f(b, y''), z''), z'), z''')), z'0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 19
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(f(f(y'', z''), z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(f(b, y''), z''), z'), z''')), z'0))
F(f(f(a, b), c), f(f(f(y'', z''), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y''), z'')), z''''), z'0))
F(f(f(a, b), c), f(f(y''''', z''''), z'0)) -> F(a, f(f(f(f(c, b), y'''''), z''''), z'0))
F(f(f(a, b), c), f(f(f(f(y', z1), z0), z'), z'')) -> F(a, f(f(c, f(f(f(f(b, y'), z1), z0), z')), z''))
F(f(f(a, b), c), f(f(f(y''', z0'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'''), z0')), z'''), z''))
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'), z')), z'''), z''))
F(f(f(a, b), c), f(f(y''', z'''), z'')) -> F(a, f(f(f(f(c, b), y'''), z'''), z''))
F(f(f(a, b), c), f(f(f(f(y'', z1), z0), z''), z')) -> F(a, f(c, f(f(f(f(f(b, y''), z1), z0), z''), z')))
F(f(f(a, b), c), f(f(f(y'', z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y''), z0'), z''')), z'0))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(f(y', z'), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y'), z')), z''''), z'0))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(f(f(y'', z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(b, y''), z0'), z''')), z'0))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(f(f(y''', z0''), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y'''), z0'')), z''''), z'0))
F(f(f(a, b), c), f(f(f(f(y', z'), z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(f(b, y'), z'), z0'), z''')), z'0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 20
Narrowing Transformation


Dependency Pairs:

F(f(f(a, b), c), f(f(f(f(y', z'), z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(f(b, y'), z'), z0'), z''')), z'0))
F(f(f(a, b), c), f(f(f(y''', z0''), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y'''), z0'')), z''''), z'0))
F(f(f(a, b), c), f(f(f(y'', z''), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y''), z'')), z''''), z'0))
F(f(f(a, b), c), f(f(f(y', z'), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y'), z')), z''''), z'0))
F(f(f(a, b), c), f(f(y''''', z''''), z'0)) -> F(a, f(f(f(f(c, b), y'''''), z''''), z'0))
F(f(f(a, b), c), f(f(f(f(y', z1), z0), z'), z'')) -> F(a, f(f(c, f(f(f(f(b, y'), z1), z0), z')), z''))
F(f(f(a, b), c), f(f(f(y''', z0'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'''), z0')), z'''), z''))
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'), z')), z'''), z''))
F(f(f(a, b), c), f(f(y''', z'''), z'')) -> F(a, f(f(f(f(c, b), y'''), z'''), z''))
F(f(f(a, b), c), f(f(f(f(y'', z1), z0), z''), z')) -> F(a, f(c, f(f(f(f(f(b, y''), z1), z0), z''), z')))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(f(f(y'', z''), z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(f(b, y''), z''), z'), z''')), z'0))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(f(f(a, b), c), f(f(f(f(y'', z1), z0), z''), z')) -> F(a, f(c, f(f(f(f(f(b, y''), z1), z0), z''), z')))
two new Dependency Pairs are created:

F(f(f(a, b), c), f(f(f(f(y''', z1'), z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(f(b, y'''), z1'), z0'), z''')), z'0))
F(f(f(a, b), c), f(f(f(f(f(y', z2), z1), z0), z''), z')) -> F(a, f(c, f(f(f(f(f(f(b, y'), z2), z1), z0), z''), z')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 21
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

F(f(f(a, b), c), f(f(f(f(f(y', z2), z1), z0), z''), z')) -> F(a, f(c, f(f(f(f(f(f(b, y'), z2), z1), z0), z''), z')))
F(f(f(a, b), c), f(f(f(f(y''', z1'), z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(f(b, y'''), z1'), z0'), z''')), z'0))
F(f(f(a, b), c), f(f(f(y''', z0''), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y'''), z0'')), z''''), z'0))
F(f(f(a, b), c), f(f(f(f(y'', z''), z'), z'''), z'0)) -> F(a, f(f(c, f(f(f(f(b, y''), z''), z'), z''')), z'0))
F(f(f(a, b), c), f(f(f(y'', z''), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y''), z'')), z''''), z'0))
F(f(f(a, b), c), f(f(f(y', z'), z''''), z'0)) -> F(a, f(f(f(c, f(f(b, y'), z')), z''''), z'0))
F(f(f(a, b), c), f(f(y''''', z''''), z'0)) -> F(a, f(f(f(f(c, b), y'''''), z''''), z'0))
F(f(f(a, b), c), f(f(f(f(y', z1), z0), z'), z'')) -> F(a, f(f(c, f(f(f(f(b, y'), z1), z0), z')), z''))
F(f(f(a, b), c), f(f(f(y''', z0'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'''), z0')), z'''), z''))
F(f(f(a, b), c), f(f(f(y', z'), z'''), z'')) -> F(a, f(f(f(c, f(f(b, y'), z')), z'''), z''))
F(f(f(a, b), c), f(f(y''', z'''), z'')) -> F(a, f(f(f(f(c, b), y'''), z'''), z''))
F(f(f(a, b), c), f(y''', z'')) -> F(a, f(f(f(c, b), y'''), z''))
F(f(f(a, b), c), f(f(f(y'''', z''''), z'0'), z)) -> F(f(f(a, b), c), f(f(y'''', z''''), z'0'))
F(f(f(a, b), c), f(f(y'''', z''''), z)) -> F(f(f(a, b), c), f(y'''', z''''))
F(f(f(a, b), c), f(f(y''', z'''), z)) -> F(f(f(a, b), c), f(y''', z'''))
F(f(f(a, b), c), f(y', z)) -> F(f(f(a, b), c), y')
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(f(a, b), c), x'') -> F(a, f(f(c, b), x''))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(f(a, b), c), f(f(f(f(y', z'), z0'), z'''), z'0)) -> F(a, f(f(c, f(f(f(f(b, y'), z'), z0'), z''')), z'0))


Rules:


f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:06 minutes