Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

W(r(x)) → W(x)
B(r(x)) → B(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.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

W(r(x)) → W(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
W(x1)  =  W(x1)
r(x1)  =  r(x1)
w(x1)  =  x1
b(x1)  =  b(x1)

Lexicographic path order with status [LPO].
Precedence:

W₁ > r₁
b₁ > r₁

Status:

W₁: [1]
r₁: [1]
b₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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.

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

B(r(x)) → B(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
B(x1)  =  B(x1)
w(x1)  =  x1
r(x1)  =  r(x1)
b(x1)  =  b(x1)

Lexicographic path order with status [LPO].
Precedence:

b₁ > r₁

Status:

B₁: [1]
r₁: [1]
b₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

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.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

B(w(x)) → B(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
B(x1)  =  x1
w(x1)  =  w(x1)
r(x1)  =  r
b(x1)  =  b(x1)

Lexicographic path order with status [LPO].
Precedence:

b₁ > w₁
b₁ > r

Status:

w₁: [1]
r: []
b₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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.

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE