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 monotone reduction pair:

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

The next rules are strictly ordered:

  p2, p3, p4
  r1

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: a#(a(x)) -> b#(x)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)