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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        b# > r > w
      
      argument filter:
    
        pi(b#) = 1
        pi(w) = [1]
        pi(r) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        b#_A(x1) = x1
        w_A(x1) = x1 + 1
        r_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3

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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        w# > r
      
      argument filter:
    
        pi(w#) = 1
        pi(r) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        w#_A(x1) = x1
        r_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1
  r1, r2, r3

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