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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = max{4, x1 - 5}
        f_A(x1) = max{11, x1 + 10}
        g_A(x1,x2) = max{x1 - 4, x2 + 9}
        h_A(x1,x2) = max{11, x1 - 3, x2}
        |0|_A = 2
  
    precedence:
    
      f# = f = g = h = |0|
    
    partial status:
  
      pi(f#) = []
      pi(f) = []
      pi(g) = []
      pi(h) = [2]
      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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = max{15, x1 + 7}
        f_A(x1) = 17
        h_A(x1,x2) = max{5, x1 - 8, x2 - 10}
        g_A(x1,x2) = max{5, x1 - 8, x2 + 4}
        |0|_A = 1
  
    precedence:
    
      f# = f = h = g = |0|
    
    partial status:
  
      pi(f#) = [1]
      pi(f) = []
      pi(h) = []
      pi(g) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p1

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