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.

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 contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

W1(r1(x)) -> W1(x)
Used argument filtering: W1(x1)  =  x1
r1(x1)  =  r1(x1)
Used ordering: Precedence:
trivial



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

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

B1(r1(x)) -> B1(x)
Used argument filtering: B1(x1)  =  x1
w1(x1)  =  x1
r1(x1)  =  r1(x1)
Used ordering: Precedence:
trivial



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

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

B1(w1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

B1(w1(x)) -> B1(x)
Used argument filtering: B1(x1)  =  x1
w1(x1)  =  w1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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.