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
                 ↳Forward Instantiation 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 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 4
                 ↳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), 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, f(f(a, c), 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 Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(x, f(y, z)) -> F(x, y)
four 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''''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 5
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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



Innermost Termination of R could not be shown.
Duration:
0:01 minutes