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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1) = x1
        a_A(x1) = x1 + 2
        b#_A(x1) = x1
        b_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1) = 2
        a_A(x1) = 2
        b#_A(x1) = x1
        b_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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