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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        c#_A(x1) = max{33, x1 + 3}
        b_A(x1) = max{34, x1 + 21}
        a_A(x1) = max{13, x1 - 18}
        c_A(x1) = max{5, x1 - 4}
        e_A = 0
  
    precedence:
    
      b > e > c > a > c#
    
    partial status:
  
      pi(c#) = [1]
      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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        c#_A(x1) = x1 + 2
        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.