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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1,x2) = ((0,1),(0,0)) x2 + (1,5)
            a_A() = (2,6)
            f_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (0,1)
            b_A() = (1,4)
    
      precedence:
      
        f# = a = f = b
      
      partial status:
    
        pi(f#) = []
        pi(a) = []
        pi(f) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1,x2) = (1,1)
            a_A() = (2,3)
            f_A(x1,x2) = (0,2)
            b_A() = (1,3)
    
      precedence:
      
        f# = a = f = b
      
      partial status:
    
        pi(f#) = []
        pi(a) = []
        pi(f) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p1

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