YES

We show the termination of the TRS R:

  f(g(x)) -> g(f(f(x)))
  f(h(x)) -> h(g(x))
  |f'|(s(x),y,y) -> |f'|(y,x,s(x))

-- SCC decomposition.

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

p1: f#(g(x)) -> f#(f(x))
p2: f#(g(x)) -> f#(x)
p3: |f'|#(s(x),y,y) -> |f'|#(y,x,s(x))

and R consists of:

r1: f(g(x)) -> g(f(f(x)))
r2: f(h(x)) -> h(g(x))
r3: |f'|(s(x),y,y) -> |f'|(y,x,s(x))

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#(g(x)) -> f#(f(x))
p2: f#(g(x)) -> f#(x)

and R consists of:

r1: f(g(x)) -> g(f(f(x)))
r2: f(h(x)) -> h(g(x))
r3: |f'|(s(x),y,y) -> |f'|(y,x,s(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:
        
          f#_A(x1) = x1 + 2
          g_A(x1) = x1 + 2
          f_A(x1) = x1
          h_A(x1) = 0
    
      precedence:
      
        f > f# = g > h
      
      partial status:
    
        pi(f#) = [1]
        pi(g) = [1]
        pi(f) = []
        pi(h) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 5
          g_A(x1) = x1 + 5
          f_A(x1) = 5
          h_A(x1) = 15
    
      precedence:
      
        g = f = h > f#
      
      partial status:
    
        pi(f#) = [1]
        pi(g) = [1]
        pi(f) = []
        pi(h) = []
    

The next rules are strictly ordered:

  p1, p2

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: |f'|#(s(x),y,y) -> |f'|#(y,x,s(x))

and R consists of:

r1: f(g(x)) -> g(f(f(x)))
r2: f(h(x)) -> h(g(x))
r3: |f'|(s(x),y,y) -> |f'|(y,x,s(x))

The set of usable rules consists of

  (no rules)

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,x3) = max{x1 + 2, x2 + 3, x3 + 1}
          s_A(x1) = max{5, x1 + 2}
    
      precedence:
      
        |f'|# = s
      
      partial status:
    
        pi(|f'|#) = [1, 2, 3]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          |f'|#_A(x1,x2,x3) = max{x1 + 1, x2 + 2, x3 + 2}
          s_A(x1) = max{3, x1}
    
      precedence:
      
        |f'|# = s
      
      partial status:
    
        pi(|f'|#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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