Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Forward Instantiation Transformation

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 9

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 10

                 ↳Polynomial Ordering

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c) = 0

POL(b) = 0

POL(a) = 0

POL(f(x₁, x₂)) = 1 + x₂

POL(F(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 11

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 12

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 13

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 14

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 15

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 16

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 17

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 18

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 19

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 20

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

POL(c)	= 0
POL(b)	= 0
POL(a)	= 0
POL(f(x₁, x₂))	= 1 + x₂
POL(F(x₁, x₂))	= x₂