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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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

B(w(x)) → W(b(x))
B(w(x)) → B(x)
W(r(x)) → W(x)
B(r(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 deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

B(w(x)) → W(b(x))
B(w(x)) → B(x)
W(r(x)) → W(x)
B(r(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 [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
W(x1)  =  W(x1)
r(x1)  =  r(x1)

Lexicographic Path Order [19].
Precedence:
r1 > W1

The following usable rules [14] were oriented: none



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

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

B(w(x)) → B(x)
B(r(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 [13].


The following pairs can be oriented strictly and are deleted.


B(w(x)) → B(x)
The remaining pairs can at least be oriented weakly.

B(r(x)) → B(x)
Used ordering: Combined order from the following AFS and order.
B(x1)  =  B(x1)
w(x1)  =  w(x1)
r(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
w1 > B1

The following usable rules [14] were oriented: none



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

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

B(r(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 [13].


The following pairs can be oriented strictly and are deleted.


B(r(x)) → B(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
B(x1)  =  B(x1)
r(x1)  =  r(x1)

Lexicographic Path Order [19].
Precedence:
r1 > B1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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.