YES

We show the termination of the TRS R:

  f(g(X)) -> g(f(f(X)))
  f(h(X)) -> h(g(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)

and R consists of:

r1: f(g(X)) -> g(f(f(X)))
r2: f(h(X)) -> h(g(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = x1 + (4,1)
            g_A(x1) = ((1,0),(0,0)) x1 + (5,3)
            f_A(x1) = ((1,0),(1,0)) x1 + (0,1)
            h_A(x1) = (4,4)
    
      precedence:
      
        f > g > f# > h
      
      partial status:
    
        pi(f#) = [1]
        pi(g) = []
        pi(f) = []
        pi(h) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = (0,0)
            g_A(x1) = (0,0)
            f_A(x1) = (2,2)
            h_A(x1) = (1,1)
    
      precedence:
      
        f# > g = f = h
      
      partial status:
    
        pi(f#) = []
        pi(g) = []
        pi(f) = []
        pi(h) = []
    

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#(g(X)) -> f#(X)

and R consists of:

r1: f(g(X)) -> g(f(f(X)))
r2: f(h(X)) -> h(g(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(g(X)) -> f#(X)

and R consists of:

r1: f(g(X)) -> g(f(f(X)))
r2: f(h(X)) -> h(g(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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            g_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        f# = g
      
      partial status:
    
        pi(f#) = []
        pi(g) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = ((1,0),(0,0)) x1
            g_A(x1) = (1,1)
    
      precedence:
      
        g > f#
      
      partial status:
    
        pi(f#) = []
        pi(g) = []
    

The next rules are strictly ordered:

  p1

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