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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            c#_A(x1) = x1 + (0,4)
            c_A(x1) = ((1,0),(0,0)) x1 + (1,3)
            a_A(x1,x2) = x1 + (0,3)
            b_A(x1,x2) = x1 + x2 + (0,1)
            |0|_A() = (0,1)
    
      precedence:
      
        |0| > c > c# = a > b
      
      partial status:
    
        pi(c#) = [1]
        pi(c) = []
        pi(a) = []
        pi(b) = [1, 2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            c#_A(x1) = ((1,0),(0,0)) x1 + (0,3)
            c_A(x1) = (3,4)
            a_A(x1,x2) = (2,2)
            b_A(x1,x2) = x1 + x2
            |0|_A() = (1,1)
    
      precedence:
      
        c# = c = |0| > a > b
      
      partial status:
    
        pi(c#) = []
        pi(c) = []
        pi(a) = []
        pi(b) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p2

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#(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#(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:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            c#_A(x1) = ((0,1),(0,0)) x1 + (1,16)
            c_A(x1) = ((0,0),(1,0)) x1 + (16,1)
            a_A(x1,x2) = ((0,1),(0,1)) x1 + x2 + (1,6)
            b_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,0),(1,1)) x2 + (1,1)
            |0|_A() = (2,1)
    
      precedence:
      
        c# > a = |0| > c > b
      
      partial status:
    
        pi(c#) = []
        pi(c) = []
        pi(a) = [2]
        pi(b) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            c#_A(x1) = (2,1)
            c_A(x1) = (3,1)
            a_A(x1,x2) = (4,2)
            b_A(x1,x2) = (1,2)
            |0|_A() = (5,3)
    
      precedence:
      
        a = b = |0| > c# = c
      
      partial status:
    
        pi(c#) = []
        pi(c) = []
        pi(a) = []
        pi(b) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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