Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A(f, a(f, a(g, a(g, x)))) -> A(g, a(g, a(g, a(f, a(f, a(f, x))))))
A(f, a(f, a(g, a(g, x)))) -> A(g, a(g, a(f, a(f, a(f, x)))))
A(f, a(f, a(g, a(g, x)))) -> A(g, a(f, a(f, a(f, x))))
A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, a(f, x)))
A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, x))
A(f, a(f, a(g, a(g, x)))) -> A(f, x)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

A(f, a(f, a(g, a(g, x)))) -> A(f, x)
A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, x))
A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, a(f, x)))

Rule:

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

Strategy:

innermost

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

A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, a(f, x)))

two new Dependency Pairs are created:

A(f, a(f, a(g, a(g, a(g, a(g, x'')))))) -> A(f, a(g, a(g, a(g, a(f, a(f, a(f, x'')))))))
A(f, a(f, a(g, a(g, a(f, a(g, a(g, x''))))))) -> A(f, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x''))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

A(f, a(f, a(g, a(g, a(f, a(g, a(g, x''))))))) -> A(f, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x''))))))))
A(f, a(f, a(g, a(g, a(g, a(g, x'')))))) -> A(f, a(g, a(g, a(g, a(f, a(f, a(f, x'')))))))
A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, x))
A(f, a(f, a(g, a(g, x)))) -> A(f, x)

Rule:

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

Strategy:

innermost

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

A(f, a(f, a(g, a(g, a(g, a(g, x'')))))) -> A(f, a(g, a(g, a(g, a(f, a(f, a(f, x'')))))))

two new Dependency Pairs are created:

A(f, a(f, a(g, a(g, a(g, a(g, a(g, a(g, x')))))))) -> A(f, a(g, a(g, a(g, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x')))))))))))
A(f, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x'))))))))) -> A(f, a(g, a(g, a(g, a(f, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x'))))))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Rewriting Transformation

Dependency Pairs:

A(f, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x'))))))))) -> A(f, a(g, a(g, a(g, a(f, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x'))))))))))))
A(f, a(f, a(g, a(g, a(g, a(g, a(g, a(g, x')))))))) -> A(f, a(g, a(g, a(g, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x')))))))))))
A(f, a(f, a(g, a(g, x)))) -> A(f, x)
A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, x))
A(f, a(f, a(g, a(g, a(f, a(g, a(g, x''))))))) -> A(f, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x''))))))))

Rule:

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

Strategy:

innermost

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

A(f, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x'))))))))) -> A(f, a(g, a(g, a(g, a(f, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x'))))))))))))

one new Dependency Pair is created:

A(f, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x'))))))))) -> A(f, a(g, a(g, a(g, a(g, a(g, a(g, a(f, a(f, a(f, a(g, a(f, a(f, a(f, x'))))))))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

A(f, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x'))))))))) -> A(f, a(g, a(g, a(g, a(g, a(g, a(g, a(f, a(f, a(f, a(g, a(f, a(f, a(f, x'))))))))))))))
A(f, a(f, a(g, a(g, a(f, a(g, a(g, x''))))))) -> A(f, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x''))))))))
A(f, a(f, a(g, a(g, x)))) -> A(f, x)
A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, x))
A(f, a(f, a(g, a(g, a(g, a(g, a(g, a(g, x')))))))) -> A(f, a(g, a(g, a(g, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x')))))))))))

Rule:

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

Strategy:

innermost

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

A(f, a(f, a(g, a(g, x)))) -> A(f, x)

four new Dependency Pairs are created:

A(f, a(f, a(g, a(g, a(f, a(g, a(g, x''))))))) -> A(f, a(f, a(g, a(g, x''))))
A(f, a(f, a(g, a(g, a(f, a(g, a(g, a(f, a(g, a(g, x'''')))))))))) -> A(f, a(f, a(g, a(g, a(f, a(g, a(g, x'''')))))))
A(f, a(f, a(g, a(g, a(f, a(g, a(g, a(g, a(g, a(g, a(g, x'''))))))))))) -> A(f, a(f, a(g, a(g, a(g, a(g, a(g, a(g, x'''))))))))
A(f, a(f, a(g, a(g, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x''')))))))))))) -> A(f, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x''')))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Polynomial Ordering

Dependency Pairs:

A(f, a(f, a(g, a(g, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x''')))))))))))) -> A(f, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x''')))))))))
A(f, a(f, a(g, a(g, a(f, a(g, a(g, a(g, a(g, a(g, a(g, x'''))))))))))) -> A(f, a(f, a(g, a(g, a(g, a(g, a(g, a(g, x'''))))))))
A(f, a(f, a(g, a(g, a(f, a(g, a(g, a(f, a(g, a(g, x'''')))))))))) -> A(f, a(f, a(g, a(g, a(f, a(g, a(g, x'''')))))))
A(f, a(f, a(g, a(g, a(f, a(g, a(g, x''))))))) -> A(f, a(f, a(g, a(g, x''))))
A(f, a(f, a(g, a(g, a(g, a(g, a(g, a(g, x')))))))) -> A(f, a(g, a(g, a(g, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x')))))))))))
A(f, a(f, a(g, a(g, a(f, a(g, a(g, x''))))))) -> A(f, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x''))))))))
A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, x))
A(f, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x'))))))))) -> A(f, a(g, a(g, a(g, a(g, a(g, a(g, a(f, a(f, a(f, a(g, a(f, a(f, a(f, x'))))))))))))))

Rule:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

A(f, a(f, a(g, a(g, a(g, a(g, a(g, a(g, x')))))))) -> A(f, a(g, a(g, a(g, a(f, a(g, a(g, a(g, a(f, a(f, a(f, x')))))))))))
A(f, a(f, a(g, a(g, a(g, a(g, a(f, a(g, a(g, x'))))))))) -> A(f, a(g, a(g, a(g, a(g, a(g, a(g, a(f, a(f, a(f, a(g, a(f, a(f, a(f, x'))))))))))))))

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(g) = 0

POL(a(x₁, x₂)) = x₁

POL(f) = 1

POL(A(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

Rule:

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

Strategy:

innermost

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

POL(g)	= 0
POL(a(x₁, x₂))	= x₁
POL(f)	= 1
POL(A(x₁, x₂))	= x₂