YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4}


-- Reduction pair.

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

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

and R consists of:

r1: a(a(y,|0|()),|0|()) -> y
r2: c(c(y)) -> y
r3: c(a(c(c(y)),x)) -> a(c(c(c(a(x,|0|())))),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) = x1 + 33
          a_A(x1,x2) = max{49, x1 + 5, x2 + 35}
          c_A(x1) = max{69, x1 + 12}
          |0|_A = 11
    
      precedence:
      
        c# > c > |0| > a
      
      partial status:
    
        pi(c#) = [1]
        pi(a) = []
        pi(c) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = max{40, x1 + 21}
          a_A(x1,x2) = 24
          c_A(x1) = 18
          |0|_A = 1
    
      precedence:
      
        c# = a = c = |0|
      
      partial status:
    
        pi(c#) = []
        pi(a) = []
        pi(c) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: c#(a(c(c(y)),x)) -> c#(a(x,|0|()))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: c#(a(c(c(y)),x)) -> c#(a(x,|0|()))

and R consists of:

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

The set of usable rules consists of

  r1

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{19, x1 + 9}
          a_A(x1,x2) = max{x1 + 8, x2 + 15}
          c_A(x1) = x1 + 5
          |0|_A = 2
    
      precedence:
      
        c# = a = c = |0|
      
      partial status:
    
        pi(c#) = [1]
        pi(a) = [1, 2]
        pi(c) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = x1 + 10
          a_A(x1,x2) = max{x1 + 2, x2 + 8}
          c_A(x1) = x1 + 3
          |0|_A = 2
    
      precedence:
      
        c# = a = c = |0|
      
      partial status:
    
        pi(c#) = []
        pi(a) = []
        pi(c) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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