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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2) = ((1,1),(1,1)) x2 + (1,0)
          a_A() = (2,5)
          f_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (2,0)
          b_A() = (1,2)
  
    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.