YES
We show the termination of the TRS R:

  w(r(x)) -> r(w(x))
  b(r(x)) -> r(b(x))
  b(w(x)) -> w(b(x))


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: w#(r(x)) -> w#(x)
p2: b#(r(x)) -> b#(x)
p3: b#(w(x)) -> w#(b(x))
p4: b#(w(x)) -> b#(x)

and R consists of:

r1: w(r(x)) -> r(w(x))
r2: b(r(x)) -> r(b(x))
r3: b(w(x)) -> w(b(x))

The estimated dependency graph contains the following SCCs:

  {p2, p4}
  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: b#(w(x)) -> b#(x)
p2: b#(r(x)) -> b#(x)

and R consists of:

r1: w(r(x)) -> r(w(x))
r2: b(r(x)) -> r(b(x))
r3: b(w(x)) -> w(b(x))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            b#_A(0) = 0
            b#_A(1) = 0
            w_A(0) = 0
            w_A(1) = 1
            r_A(0) = 0
            r_A(1) = 1
            b_A(0) = 0
            b_A(1) = 0
            b#!_A(0) = 0
            b#!_A(1) = 0
            w!_A(0) = 0
            w!_A(1) = 1
            r!_A(0) = 0
            r!_A(1) = 1
            b!_A(0) = 1
            b!_A(1) = 1
  
    partial status:
  
      pi(b#) = [1]
      pi(w) = [1]
      pi(r) = [1]
      pi(b) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: w#(r(x)) -> w#(x)

and R consists of:

r1: w(r(x)) -> r(w(x))
r2: b(r(x)) -> r(b(x))
r3: b(w(x)) -> w(b(x))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            w#_A(0) = 0
            w#_A(1) = 0
            r_A(0) = 0
            r_A(1) = 0
            w_A(0) = 0
            w_A(1) = 0
            b_A(0) = 0
            b_A(1) = 1
            w#!_A(0) = 0
            w#!_A(1) = 0
            r!_A(0) = 0
            r!_A(1) = 0
            w!_A(0) = 0
            w!_A(1) = 0
            b!_A(0) = 0
            b!_A(1) = 0
  
    partial status:
  
      pi(w#) = [1]
      pi(r) = [1]
      pi(w) = [1]
      pi(b) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.