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:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1) = ((1,0),(0,0)) x1 + (1,2)
          f_A(x1) = x1 + (4,4)
          g_A(x1,x2) = x2 + (0,1)
          h_A(x1,x2) = (2,3)
          |0|_A() = (1,1)
  
    precedence:
    
      f = g > f# = h > |0|
    
    partial status:
  
      pi(f#) = []
      pi(f) = [1]
      pi(g) = []
      pi(h) = []
      pi(|0|) = []

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(x)) -> f#(g(f(x),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#(g(f(x),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:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1) = ((1,0),(0,0)) x1 + (4,9)
          f_A(x1) = ((1,0),(1,0)) x1 + (4,3)
          g_A(x1,x2) = ((1,0),(1,1)) x2 + (1,8)
          h_A(x1,x2) = (3,2)
          |0|_A() = (1,1)
  
    precedence:
    
      f# = f = g = h = |0|
    
    partial status:
  
      pi(f#) = []
      pi(f) = []
      pi(g) = [2]
      pi(h) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p1

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