YES

We show the termination of the TRS R:

  a(b(x)) -> b(b(a(x)))

-- SCC decomposition.

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

p1: a#(b(x)) -> a#(x)

and R consists of:

r1: a(b(x)) -> b(b(a(x)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: a#(b(x)) -> a#(x)

and R consists of:

r1: a(b(x)) -> b(b(a(x)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      a#_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1
      b_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)

The next rules are strictly ordered:

  p1
  r1

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