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. max/plus interpretations on natural numbers:
    
      c#_A(x1,x2,x3) = max{11, x1 - 63, x2 - 76, x3 - 110}
      a_A = 137
      b#_A(x1,x2) = max{0, x1 - 75, x2 - 77}
      b_A(x1,x2) = max{x1 + 89, x2 - 46}
      f_A(x1) = max{12, x1 - 127}
      c_A(x1,x2,x3) = max{82, x1 + 64, x2 - 1, x3 + 81}
    
    2. max/plus interpretations on natural numbers:
    
      c#_A(x1,x2,x3) = 1
      a_A = 0
      b#_A(x1,x2) = 0
      b_A(x1,x2) = 2
      f_A(x1) = 1
      c_A(x1,x2,x3) = 0
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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