YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = x1 + 10
          a_A = 4
          f_A(x1,x2) = 3
    
      precedence:
      
        f# = a > f
      
      partial status:
    
        pi(f#) = [1]
        pi(a) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{24, x1 + 17}
          a_A = 13
          f_A(x1,x2) = 0
    
      precedence:
      
        f# = a = f
      
      partial status:
    
        pi(f#) = []
        pi(a) = []
        pi(f) = []
    

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

and R consists of:

r1: f(a(),f(x,a())) -> f(a(),f(f(a(),x),f(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#(a(),f(x,a())) -> f#(a(),f(f(a(),x),f(a(),a())))
p2: f#(a(),f(x,a())) -> f#(a(),x)

and R consists of:

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

The set of usable rules consists of

  r1

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 14, x2 + 25}
          a_A = 17
          f_A(x1,x2) = max{0, x2 - 6}
    
      precedence:
      
        f# = a = f
      
      partial status:
    
        pi(f#) = [2]
        pi(a) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = x2 + 23
          a_A = 30
          f_A(x1,x2) = 17
    
      precedence:
      
        f# = a = f
      
      partial status:
    
        pi(f#) = []
        pi(a) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1

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