Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(f(f(a, f(a, a)), a), x) → f(x, f(x, a))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(f(f(a, f(a, a)), a), x) → f(x, f(x, a))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(f(f(a, f(a, a)), a), x) → F(x, f(x, a))
F(f(f(a, f(a, a)), a), x) → F(x, a)

The TRS R consists of the following rules:

f(f(f(a, f(a, a)), a), x) → f(x, f(x, a))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

F(f(f(a, f(a, a)), a), x) → F(x, f(x, a))
F(f(f(a, f(a, a)), a), x) → F(x, a)

The TRS R consists of the following rules:

f(f(f(a, f(a, a)), a), x) → f(x, f(x, a))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(f(f(a, f(a, a)), a), x) → F(x, a)
The remaining pairs can at least be oriented weakly.

F(f(f(a, f(a, a)), a), x) → F(x, f(x, a))
Used ordering: Polynomial interpretation [25,35]:

POL(a) = 0   
POL(f(x1, x2)) = 1/4   
POL(F(x1, x2)) = (1/4)x_1 + (1/4)x_2   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

f(f(f(a, f(a, a)), a), x) → f(x, f(x, a))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

F(f(f(a, f(a, a)), a), x) → F(x, f(x, a))

The TRS R consists of the following rules:

f(f(f(a, f(a, a)), a), x) → f(x, f(x, a))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.