YES

We show the termination of the TRS R:

  b(b(|0|(),y),x) -> y
  c(c(c(y))) -> c(c(a(a(c(b(|0|(),y)),|0|()),|0|())))
  a(y,|0|()) -> b(y,|0|())

-- SCC decomposition.

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

p1: c#(c(c(y))) -> c#(c(a(a(c(b(|0|(),y)),|0|()),|0|())))
p2: c#(c(c(y))) -> c#(a(a(c(b(|0|(),y)),|0|()),|0|()))
p3: c#(c(c(y))) -> a#(a(c(b(|0|(),y)),|0|()),|0|())
p4: c#(c(c(y))) -> a#(c(b(|0|(),y)),|0|())
p5: c#(c(c(y))) -> c#(b(|0|(),y))
p6: c#(c(c(y))) -> b#(|0|(),y)
p7: a#(y,|0|()) -> b#(y,|0|())

and R consists of:

r1: b(b(|0|(),y),x) -> y
r2: c(c(c(y))) -> c(c(a(a(c(b(|0|(),y)),|0|()),|0|())))
r3: a(y,|0|()) -> b(y,|0|())

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: c#(c(c(y))) -> c#(c(a(a(c(b(|0|(),y)),|0|()),|0|())))
p2: c#(c(c(y))) -> c#(a(a(c(b(|0|(),y)),|0|()),|0|()))

and R consists of:

r1: b(b(|0|(),y),x) -> y
r2: c(c(c(y))) -> c(c(a(a(c(b(|0|(),y)),|0|()),|0|())))
r3: a(y,|0|()) -> b(y,|0|())

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) = max{0, x1 - 39}
      c_A(x1) = x1 + 28
      a_A(x1,x2) = max{x1 - 5, x2 + 10}
      b_A(x1,x2) = max{9, x1 - 6, x2 + 7}
      |0|_A = 0
    
    2. max/plus interpretations on natural numbers:
    
      c#_A(x1) = 0
      c_A(x1) = x1 + 2
      a_A(x1,x2) = 3
      b_A(x1,x2) = 4
      |0|_A = 0
    

The next rules are strictly ordered:

  p1, p2

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