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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        f#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x2
        a_A() = (2,1,1,1)
        f_A(x1,x2) = x1 + ((0,0,0,0),(1,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + (0,1,1,1)
        b_A() = (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        b > a > f > f#
      
      argument filter:
    
        pi(f#) = 1
        pi(a) = []
        pi(f) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p1

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