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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          b_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      a# = b
    
    partial status:
  
      pi(a#) = []
      pi(b) = []

The next rules are strictly ordered:

  p1

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