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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      c#_A(x1) = x1
      b_A(x1) = ((1,1),(1,0)) x1
      a_A(x1) = ((0,0),(1,0)) x1 + (1,1)
      c_A(x1) = ((0,0),(1,1)) x1 + (2,1)
      e_A() = (0,0)

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

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      c#_A(x1) = ((0,1),(0,0)) x1
      b_A(x1) = ((0,1),(0,1)) x1
      a_A(x1) = ((0,0),(1,1)) x1 + (0,1)
      c_A(x1) = ((0,0),(1,1)) x1
      e_A() = (0,0)

The next rules are strictly ordered:

  p1

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