YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

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

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

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

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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