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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1) = max{11, x1 + 5}
          b_A(x1) = max{3, x1 + 2}
          a_A(x1) = x1 + 4
    
      precedence:
      
        a# = b = a
      
      partial status:
    
        pi(a#) = [1]
        pi(b) = [1]
        pi(a) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1) = max{8, x1 - 2}
          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.