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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        b#_A(x1) = ((1,1),(1,1)) x1
        w_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        r_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        b#_A(x1) = ((0,1),(1,1)) x1
        w_A(x1) = x1 + (1,1)
        r_A(x1) = ((0,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        w#_A(x1) = ((1,1),(1,1)) x1
        r_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        w#_A(x1) = ((0,1),(1,1)) x1
        r_A(x1) = ((1,0),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.