YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: a#(a(x)) -> b#(b(x))
p2: a#(a(x)) -> b#(x)
p3: b#(b(a(x))) -> a#(b(b(x)))
p4: b#(b(a(x))) -> b#(b(x))
p5: b#(b(a(x))) -> b#(x)

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: a#(a(x)) -> b#(b(x))
p2: b#(b(a(x))) -> b#(x)
p3: b#(b(a(x))) -> b#(b(x))
p4: b#(b(a(x))) -> a#(b(b(x)))
p5: a#(a(x)) -> b#(x)

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The set of usable rules consists of

  r1, r2

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) = x1 + (2,4)
            a_A(x1) = x1 + (7,0)
            b#_A(x1) = x1 + (6,2)
            b_A(x1) = ((1,0),(0,0)) x1 + (2,1)
    
      precedence:
      
        b > a# = a > b#
      
      partial status:
    
        pi(a#) = []
        pi(a) = [1]
        pi(b#) = [1]
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = (1,1)
            a_A(x1) = (2,0)
            b#_A(x1) = (1,1)
            b_A(x1) = (2,0)
    
      precedence:
      
        a# = b# = b > a
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(b#) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(a(x)) -> b#(b(x))
p2: b#(b(a(x))) -> b#(x)
p3: b#(b(a(x))) -> a#(b(b(x)))
p4: a#(a(x)) -> b#(x)

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: a#(a(x)) -> b#(b(x))
p2: b#(b(a(x))) -> a#(b(b(x)))
p3: a#(a(x)) -> b#(x)
p4: b#(b(a(x))) -> b#(x)

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The set of usable rules consists of

  r1, r2

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

The next rules are strictly ordered:

  p1, p2, p3, p4

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