YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1) = max{3, x1 - 3}
        b_A(x1) = x1 + 6
        a_A(x1) = x1 + 1
  
    precedence:
    
      a# = b = a
    
    partial status:
  
      pi(a#) = []
      pi(b) = [1]
      pi(a) = [1]

The next rules are strictly ordered:

  p1

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