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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1,x2,x3) = ((1,1),(0,0)) x1 + ((0,1),(0,0)) x2 + ((0,1),(0,0)) x3
        a_A() = (1,6)
        b#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (1,0)
        b_A(x1,x2) = x1 + (1,3)
        f_A(x1) = ((1,0),(1,0)) x1 + (0,1)
        c_A(x1,x2,x3) = ((1,1),(0,0)) x1 + x2 + ((1,1),(0,0)) x3 + (0,2)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1,x2,x3) = (1,0)
        a_A() = (1,1)
        b#_A(x1,x2) = (0,1)
        b_A(x1,x2) = (2,2)
        f_A(x1) = (1,2)
        c_A(x1,x2,x3) = ((0,1),(0,0)) x3 + (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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