YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),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:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1,x2) = max{14, x1 + 13, x2 + 3}
          f_A = 9
          a_A(x1,x2) = max{18, x1 + 1}
          g_A = 9
    
      precedence:
      
        a# = a > f = g
      
      partial status:
    
        pi(a#) = [1, 2]
        pi(f) = []
        pi(a) = [1]
        pi(g) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1,x2) = max{22, x1 + 9, x2 - 13}
          f_A = 0
          a_A(x1,x2) = max{34, x1 + 23}
          g_A = 23
    
      precedence:
      
        a# = a = g > f
      
      partial status:
    
        pi(a#) = []
        pi(f) = []
        pi(a) = []
        pi(g) = []
    

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),x))))

The estimated dependency graph contains the following SCCs:

  {p1}
  {p3}


-- Reduction pair.

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

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),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:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1,x2) = max{31, x1 + 15, x2}
          f_A = 11
          a_A(x1,x2) = max{x1 + 24, x2 - 4}
          g_A = 1
    
      precedence:
      
        f = a = g > a#
      
      partial status:
    
        pi(a#) = [1, 2]
        pi(f) = []
        pi(a) = []
        pi(g) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1,x2) = max{15, x1 - 54}
          f_A = 70
          a_A(x1,x2) = 18
          g_A = 50
    
      precedence:
      
        a# = f = a = g
      
      partial status:
    
        pi(a#) = []
        pi(f) = []
        pi(a) = []
        pi(g) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),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:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1,x2) = max{28, x1 + 9, x2 + 19}
          g_A = 31
          a_A(x1,x2) = max{0, x1 - 7, x2 - 3}
          f_A = 20
    
      precedence:
      
        a# = g = f > a
      
      partial status:
    
        pi(a#) = [2]
        pi(g) = []
        pi(a) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1,x2) = max{48, x2 + 15}
          g_A = 58
          a_A(x1,x2) = 45
          f_A = 59
    
      precedence:
      
        a# = g = a = f
      
      partial status:
    
        pi(a#) = [2]
        pi(g) = []
        pi(a) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1

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