YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: c#(z,x,a()) -> f#(b(b(f(z),z),x))
p2: c#(z,x,a()) -> b#(b(f(z),z),x)
p3: c#(z,x,a()) -> b#(f(z),z)
p4: c#(z,x,a()) -> f#(z)
p5: b#(y,b(z,a())) -> f#(b(c(f(a()),y,z),z))
p6: b#(y,b(z,a())) -> b#(c(f(a()),y,z),z)
p7: b#(y,b(z,a())) -> c#(f(a()),y,z)
p8: b#(y,b(z,a())) -> f#(a())

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p3, p6, p7}


-- Reduction pair.

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

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

and R consists of:

r1: c(z,x,a()) -> f(b(b(f(z),z),x))
r2: b(y,b(z,a())) -> f(b(c(f(a()),y,z),z))
r3: f(c(c(z,a(),a()),x,a())) -> 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:
        
          c#_A(x1,x2,x3) = max{169, x1 + 130, x2 + 97, x3}
          a_A = 83
          b#_A(x1,x2) = max{x1 + 98, x2 + 97}
          b_A(x1,x2) = max{70, x1 + 21, x2 + 16}
          f_A(x1) = max{3, x1 - 47}
          c_A(x1,x2,x3) = max{95, x1 + 61, x2 - 1, x3 + 9}
    
      precedence:
      
        c# > a = b > f = c > b#
      
      partial status:
    
        pi(c#) = [1, 2, 3]
        pi(a) = []
        pi(b#) = [1, 2]
        pi(b) = [1, 2]
        pi(f) = []
        pi(c) = [1, 3]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1,x2,x3) = max{25, x1 + 14, x2 + 23, x3 + 1}
          a_A = 13
          b#_A(x1,x2) = max{x1 - 9, x2 + 23}
          b_A(x1,x2) = max{26, x1 + 7, x2 - 8}
          f_A(x1) = 24
          c_A(x1,x2,x3) = max{x1 + 27, x3 + 1}
    
      precedence:
      
        c# = a = b# = b = f = c
      
      partial status:
    
        pi(c#) = [3]
        pi(a) = []
        pi(b#) = []
        pi(b) = [1]
        pi(f) = []
        pi(c) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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