YES

We show the termination of the TRS R:

  c(b(a(X))) -> a(a(b(b(c(c(X))))))
  a(X) -> e()
  b(X) -> e()
  c(X) -> e()

-- SCC decomposition.

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

p1: c#(b(a(X))) -> a#(a(b(b(c(c(X))))))
p2: c#(b(a(X))) -> a#(b(b(c(c(X)))))
p3: c#(b(a(X))) -> b#(b(c(c(X))))
p4: c#(b(a(X))) -> b#(c(c(X)))
p5: c#(b(a(X))) -> c#(c(X))
p6: c#(b(a(X))) -> c#(X)

and R consists of:

r1: c(b(a(X))) -> a(a(b(b(c(c(X))))))
r2: a(X) -> e()
r3: b(X) -> e()
r4: c(X) -> e()

The estimated dependency graph contains the following SCCs:

  {p5, p6}


-- Reduction pair.

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

p1: c#(b(a(X))) -> c#(c(X))
p2: c#(b(a(X))) -> c#(X)

and R consists of:

r1: c(b(a(X))) -> a(a(b(b(c(c(X))))))
r2: a(X) -> e()
r3: b(X) -> e()
r4: c(X) -> e()

The set of usable rules consists of

  r1, r2, r3, r4

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{0, x1 - 5}
          b_A(x1) = max{6, x1 + 5}
          a_A(x1) = max{3, x1 - 5}
          c_A(x1) = 3
          e_A = 2
    
      precedence:
      
        b > e > c > c# = a
      
      partial status:
    
        pi(c#) = []
        pi(b) = []
        pi(a) = []
        pi(c) = []
        pi(e) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = 5
          b_A(x1) = 6
          a_A(x1) = 5
          c_A(x1) = 7
          e_A = 4
    
      precedence:
      
        b = c > a > c# = e
      
      partial status:
    
        pi(c#) = []
        pi(b) = []
        pi(a) = []
        pi(c) = []
        pi(e) = []
    

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#(b(a(X))) -> c#(X)

and R consists of:

r1: c(b(a(X))) -> a(a(b(b(c(c(X))))))
r2: a(X) -> e()
r3: b(X) -> e()
r4: c(X) -> e()

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: c#(b(a(X))) -> c#(X)

and R consists of:

r1: c(b(a(X))) -> a(a(b(b(c(c(X))))))
r2: a(X) -> e()
r3: b(X) -> e()
r4: c(X) -> e()

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = x1 + 3
          b_A(x1) = x1 + 1
          a_A(x1) = x1 + 1
    
      precedence:
      
        c# = b = a
      
      partial status:
    
        pi(c#) = []
        pi(b) = []
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = x1 + 1
          b_A(x1) = x1
          a_A(x1) = x1
    
      precedence:
      
        c# = b = a
      
      partial status:
    
        pi(c#) = [1]
        pi(b) = [1]
        pi(a) = [1]
    

The next rules are strictly ordered:

  p1

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