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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = max{117, x1 + 7}
          c_A(x1) = max{176, x1 + 110}
          a_A(x1,x2) = max{170, x1 - 30, x2 + 55}
          b_A(x1,x2) = max{118, x1 - 30, x2 + 54}
          |0|_A = 47
    
      precedence:
      
        c# = c > a > b = |0|
      
      partial status:
    
        pi(c#) = [1]
        pi(c) = [1]
        pi(a) = [2]
        pi(b) = [2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = max{1, x1 - 32}
          c_A(x1) = max{84, x1 + 28}
          a_A(x1,x2) = max{59, x2 + 24}
          b_A(x1,x2) = max{83, x2 + 25}
          |0|_A = 23
    
      precedence:
      
        c# = c = a = b = |0|
      
      partial status:
    
        pi(c#) = []
        pi(c) = [1]
        pi(a) = [2]
        pi(b) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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