YES

We show the termination of the TRS R:

  f(f(x)) -> f(g(f(x),x))
  f(f(x)) -> f(h(f(x),f(x)))
  g(x,y) -> y
  h(x,x) -> g(x,|0|())

-- SCC decomposition.

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

p1: f#(f(x)) -> f#(g(f(x),x))
p2: f#(f(x)) -> g#(f(x),x)
p3: f#(f(x)) -> f#(h(f(x),f(x)))
p4: f#(f(x)) -> h#(f(x),f(x))
p5: h#(x,x) -> g#(x,|0|())

and R consists of:

r1: f(f(x)) -> f(g(f(x),x))
r2: f(f(x)) -> f(h(f(x),f(x)))
r3: g(x,y) -> y
r4: h(x,x) -> g(x,|0|())

The estimated dependency graph contains the following SCCs:

  {p1, p3}


-- Reduction pair.

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

p1: f#(f(x)) -> f#(g(f(x),x))
p2: f#(f(x)) -> f#(h(f(x),f(x)))

and R consists of:

r1: f(f(x)) -> f(g(f(x),x))
r2: f(f(x)) -> f(h(f(x),f(x)))
r3: g(x,y) -> y
r4: h(x,x) -> g(x,|0|())

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{1, x1 - 6}
          f_A(x1) = max{13, x1 + 9}
          g_A(x1,x2) = max{x1 - 5, x2 + 3}
          h_A(x1,x2) = max{3, x1, x2}
          |0|_A = 0
    
      precedence:
      
        h > g > f# = f = |0|
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(g) = [2]
        pi(h) = [1, 2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = 13
          f_A(x1) = 26
          g_A(x1,x2) = 19
          h_A(x1,x2) = max{19, x2 - 9}
          |0|_A = 21
    
      precedence:
      
        |0| > h > f > f# = g
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(g) = []
        pi(h) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: f(f(x)) -> f(g(f(x),x))
r2: f(f(x)) -> f(h(f(x),f(x)))
r3: g(x,y) -> y
r4: h(x,x) -> g(x,|0|())

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

and R consists of:

r1: f(f(x)) -> f(g(f(x),x))
r2: f(f(x)) -> f(h(f(x),f(x)))
r3: g(x,y) -> y
r4: h(x,x) -> g(x,|0|())

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{0, x1 - 3}
          f_A(x1) = max{6, x1}
          h_A(x1,x2) = max{2, x2 - 10}
          g_A(x1,x2) = x2
          |0|_A = 2
    
      precedence:
      
        f = h > g = |0| > f#
      
      partial status:
    
        pi(f#) = []
        pi(f) = [1]
        pi(h) = []
        pi(g) = [2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = 19
          f_A(x1) = max{18, x1 + 7}
          h_A(x1,x2) = 17
          g_A(x1,x2) = x2 + 15
          |0|_A = 1
    
      precedence:
      
        f# = f > h > g = |0|
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(h) = []
        pi(g) = [2]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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