YES

We show the termination of the TRS R:

  b(b(y,z),c(a(),a(),a())) -> f(c(z,y,z))
  f(b(b(a(),z),c(a(),x,y))) -> z
  c(y,x,f(z)) -> b(f(b(z,x)),z)

-- SCC decomposition.

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

p1: b#(b(y,z),c(a(),a(),a())) -> f#(c(z,y,z))
p2: b#(b(y,z),c(a(),a(),a())) -> c#(z,y,z)
p3: c#(y,x,f(z)) -> b#(f(b(z,x)),z)
p4: c#(y,x,f(z)) -> f#(b(z,x))
p5: c#(y,x,f(z)) -> b#(z,x)

and R consists of:

r1: b(b(y,z),c(a(),a(),a())) -> f(c(z,y,z))
r2: f(b(b(a(),z),c(a(),x,y))) -> z
r3: c(y,x,f(z)) -> b(f(b(z,x)),z)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p5}


-- Reduction pair.

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

p1: b#(b(y,z),c(a(),a(),a())) -> c#(z,y,z)
p2: c#(y,x,f(z)) -> b#(z,x)
p3: c#(y,x,f(z)) -> b#(f(b(z,x)),z)

and R consists of:

r1: b(b(y,z),c(a(),a(),a())) -> f(c(z,y,z))
r2: f(b(b(a(),z),c(a(),x,y))) -> z
r3: c(y,x,f(z)) -> b(f(b(z,x)),z)

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          b#_A(x1,x2) = max{4, x1, x2 - 9}
          b_A(x1,x2) = max{x1 + 5, x2 + 7}
          c_A(x1,x2,x3) = max{x1 + 14, x2 + 11, x3 + 14}
          a_A = 0
          c#_A(x1,x2,x3) = max{x1 + 7, x2 + 5, x3 + 7}
          f_A(x1) = max{0, x1 - 2}
    
      precedence:
      
        c > b = a > c# > b# = f
      
      partial status:
    
        pi(b#) = [1]
        pi(b) = [1, 2]
        pi(c) = []
        pi(a) = []
        pi(c#) = [1, 2, 3]
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          b#_A(x1,x2) = x1 + 2
          b_A(x1,x2) = max{39, x1 + 20, x2 + 1}
          c_A(x1,x2,x3) = 49
          a_A = 0
          c#_A(x1,x2,x3) = max{x1 + 1, x2 + 20}
          f_A(x1) = 17
    
      precedence:
      
        c > b = a > c# = f > b#
      
      partial status:
    
        pi(b#) = [1]
        pi(b) = [2]
        pi(c) = []
        pi(a) = []
        pi(c#) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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