YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p7}


-- Reduction pair.

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

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

and R consists of:

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

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{36, x1 + 28}
          c_A(x1) = x1 + 6
          a_A(x1,x2) = max{x1 - 1, x2 + 6}
          |0|_A = 2
          b_A(x1,x2) = max{24, x1, x2 + 17}
    
      precedence:
      
        c# = c = a = |0| = b
      
      partial status:
    
        pi(c#) = [1]
        pi(c) = [1]
        pi(a) = [2]
        pi(|0|) = []
        pi(b) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = max{4, x1 + 2}
          c_A(x1) = max{0, x1 - 7}
          a_A(x1,x2) = max{19, x2 + 3}
          |0|_A = 0
          b_A(x1,x2) = x2 + 53
    
      precedence:
      
        c# = a > c = |0| = b
      
      partial status:
    
        pi(c#) = [1]
        pi(c) = []
        pi(a) = [2]
        pi(|0|) = []
        pi(b) = [2]
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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