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(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))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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(a, f(b, x)) → F(b, f(a, x))
F(a, x) → F(b, f(c, x))
F(d, f(c, x)) → F(a, x)
F(a, f(c, x)) → F(c, f(a, x))
F(a, x) → F(c, x)
F(d, f(c, x)) → F(d, f(a, x))
F(a, f(b, x)) → F(a, x)
F(a, f(c, x)) → F(a, x)

The TRS R consists of the following 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))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

F(a, f(b, x)) → F(b, f(a, x))
F(a, x) → F(b, f(c, x))
F(d, f(c, x)) → F(a, x)
F(a, f(c, x)) → F(c, f(a, x))
F(a, x) → F(c, x)
F(d, f(c, x)) → F(d, f(a, x))
F(a, f(b, x)) → F(a, x)
F(a, f(c, x)) → F(a, x)

The TRS R consists of the following 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

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

The TRS R consists of the following 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))

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(a, f(b, x)) → F(a, x)
F(a, f(c, x)) → F(a, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(c) = 7/2   
POL(a) = 0   
POL(f(x1, x2)) = 7/2 + (1/2)x_1 + (5/4)x_2   
POL(b) = 1/4   
POL(F(x1, x2)) = (1/2)x_2   
The value of delta used in the strict ordering is 29/16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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

F(d, f(c, x)) → F(d, f(a, x))

The TRS R consists of the following 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))

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