YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(a(),f(b(),x)) -> f#(a(),f(a(),f(a(),x)))
p2: f#(a(),f(b(),x)) -> f#(a(),f(a(),x))
p3: f#(a(),f(b(),x)) -> f#(a(),x)
p4: f#(b(),f(a(),x)) -> f#(b(),f(b(),f(b(),x)))
p5: f#(b(),f(a(),x)) -> f#(b(),f(b(),x))
p6: f#(b(),f(a(),x)) -> f#(b(),x)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}
  {p4, p5, p6}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      b > a > f > f#
    
    argument filter:
  
      pi(f#) = [1, 2]
      pi(a) = []
      pi(f) = [1, 2]
      pi(b) = []

The next rules are strictly ordered:

  p1, p2, p3
  r1, r2

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

-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r2

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      a > b > f# > f
    
    argument filter:
  
      pi(f#) = 2
      pi(b) = []
      pi(f) = [1, 2]
      pi(a) = []

The next rules are strictly ordered:

  p1, p2, p3

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