YES

We show the termination of the TRS R:

  f(a(),f(b(),f(a(),x))) -> f(a(),f(b(),f(b(),f(a(),x))))
  f(b(),f(b(),f(b(),x))) -> f(b(),f(b(),x))

-- SCC decomposition.

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

p1: f#(a(),f(b(),f(a(),x))) -> f#(a(),f(b(),f(b(),f(a(),x))))
p2: f#(a(),f(b(),f(a(),x))) -> f#(b(),f(b(),f(a(),x)))

and R consists of:

r1: f(a(),f(b(),f(a(),x))) -> f(a(),f(b(),f(b(),f(a(),x))))
r2: f(b(),f(b(),f(b(),x))) -> f(b(),f(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: f#(a(),f(b(),f(a(),x))) -> f#(a(),f(b(),f(b(),f(a(),x))))

and R consists of:

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

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1 + x2
      a_A() = (1,1,1)
      f_A(x1,x2) = ((1,0,0),(1,0,0),(0,0,0)) x1 + ((1,0,0),(0,0,0),(0,1,0)) x2 + (0,1,1)
      b_A() = (0,1,1)

The next rules are strictly ordered:

  p1

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