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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        c#_A(x1) = max{1, x1 - 47}
        c_A(x1) = max{86, x1 + 20}
        a_A(x1,x2) = max{x1 - 48, x2 + 62}
        b_A(x1,x2) = max{x1 - 48, x2 + 61}
        |0|_A = 3
  
    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

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        c#_A(x1) = max{77, x1 - 28}
        c_A(x1) = x1 + 58
        a_A(x1,x2) = max{x1 + 13, x2 + 81}
        b_A(x1,x2) = max{x1 - 1, x2 + 31}
        |0|_A = 12
  
    precedence:
    
      c# = c = a = b = |0|
    
    partial status:
  
      pi(c#) = []
      pi(c) = [1]
      pi(a) = []
      pi(b) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p1

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