Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(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:

B(r(x)) → B(x)
W(r(x)) → W(x)
B(w(x)) → W(b(x))
B(w(x)) → B(x)

The TRS R consists of the following rules:

w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(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:

B(r(x)) → B(x)
W(r(x)) → W(x)
B(w(x)) → W(b(x))
B(w(x)) → B(x)

The TRS R consists of the following rules:

w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(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 1 less node.

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

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

W(r(x)) → W(x)

The TRS R consists of the following rules:

w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(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.


W(r(x)) → W(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(r(x1)) = 1 + (4)x_1   
POL(W(x1)) = (4)x_1   
The value of delta used in the strict ordering is 4.
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:

w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(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
            ↳ QDPOrderProof

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

B(r(x)) → B(x)
B(w(x)) → B(x)

The TRS R consists of the following rules:

w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(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.


B(r(x)) → B(x)
B(w(x)) → B(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(w(x1)) = 4 + (4)x_1   
POL(B(x1)) = (4)x_1   
POL(r(x1)) = 4 + x_1   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



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

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

w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(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.