YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p5}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      f#_A(x1) = ((0,1),(1,1)) x1
      b_A(x1,x2) = ((0,1),(1,1)) x1 + ((1,1),(1,1)) x2
      f_A(x1) = ((1,1),(1,1)) x1 + (1,0)
      a_A() = (0,0)

The next rules are strictly ordered:

  p1

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