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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            b_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        a# = b
      
      partial status:
    
        pi(a#) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = ((1,0),(0,0)) x1
            b_A(x1) = (1,1)
    
      precedence:
      
        b > a#
      
      partial status:
    
        pi(a#) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p1

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