Termination Proof using AProVE

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))

Termination of R to be shown.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

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))

The following dependency pairs can be strictly oriented:

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

There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c) = 0

POL(b) = 0

POL(a) = 0

POL(F(x₁, x₂)) = x₁ + x₂

POL(f(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂) -> F(x₁, x₂)
f(x₁, x₂) -> f(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

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))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pair:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Narrowing Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 5

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 6

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

Dependency Pairs:

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

The following dependency pairs can be strictly oriented:

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

The following usable rules using the Ce-refinement 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) = 0

POL(a) = 1

POL(F(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂) -> F(x₁, x₂)
f(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 9

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 10

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 11

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 12

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 13

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 14

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 15

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 16

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 17

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 18

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 19

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 20

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 21

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 22

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 23

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 24

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 25

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 26

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 27

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 28

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

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

Termination of R could not be shown.
Duration:
0:12 minutes

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

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