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

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          b#_A(x1) = x1
          w_A(x1) = x1 + 2
          r_A(x1) = x1 + 3
  
    base order 2:
  
      precedence:
      
        b# > w > r
      
  
    partial status:
  
      pi(b#) = [1]
      pi(w) = [1]
      pi(r) = [1]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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

Take the monotone reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          b#_A(x1) = x1
          r_A(x1) = x1 + 3
          w_A(x1) = x1 + 2
  
    base order 2:
  
      precedence:
      
        r = w > b#
      
  
    partial status:
  
      pi(b#) = [1]
      pi(r) = [1]
      pi(w) = [1]

The following rules are removed:

  p1

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

Take the monotone reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          w#_A(x1) = x1
          r_A(x1) = x1 + 3
          w_A(x1) = x1 + 2
  
    base order 2:
  
      precedence:
      
        r = w > w#
      
  
    partial status:
  
      pi(w#) = [1]
      pi(r) = [1]
      pi(w) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.