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:

w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(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:

B1(w1(x)) -> W1(b1(x))
B1(w1(x)) -> B1(x)
W1(r1(x)) -> W1(x)
B1(r1(x)) -> B1(x)

The TRS R consists of the following rules:

w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(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:

B1(w1(x)) -> W1(b1(x))
B1(w1(x)) -> B1(x)
W1(r1(x)) -> W1(x)
B1(r1(x)) -> B1(x)

The TRS R consists of the following rules:

w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

W1(r1(x)) -> W1(x)

The TRS R consists of the following rules:

w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(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.


W1(r1(x)) -> W1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(W1(x1)) = 2·x12   
POL(r1(x1)) = 2 + 2·x12   

The following usable rules [14] 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:

w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(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:

B1(w1(x)) -> B1(x)
B1(r1(x)) -> B1(x)

The TRS R consists of the following rules:

w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(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.


B1(w1(x)) -> B1(x)
B1(r1(x)) -> B1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(B1(x1)) = 2·x12   
POL(r1(x1)) = 2 + 2·x12   
POL(w1(x1)) = 2 + 2·x12   

The following usable rules [14] 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:

w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(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.