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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1) = ((0,1),(0,0)) x1
        c_A(x1) = ((1,1),(1,1)) x1 + (13,1)
        a_A(x1,x2) = ((0,1),(0,1)) x1 + x2 + (1,1)
        b_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,0),(1,1)) x2
        |0|_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1) = (0,0)
        c_A(x1) = ((1,1),(0,0)) x1 + (3,1)
        a_A(x1,x2) = (2,1)
        b_A(x1,x2) = (1,2)
        |0|_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1) = (0,0)
        c_A(x1) = (1,1)
        a_A(x1,x2) = (2,1)
        b_A(x1,x2) = (1,1)
        |0|_A() = (1,1)
    

The next rules are strictly ordered:

  p1, p2

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