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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1) = 0
            g_A(x1) = 0
            f_A(x1) = 0
            h_A(x1) = 0
            f#!_A(x1) = 0
            g!_A(x1) = 2
            f!_A(x1) = max{1, x1 - 2}
            h!_A(x1) = max{0, x1 - 3}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1) = 0
              g_A(x1) = x1 + 1
              f_A(x1) = x1
              h_A(x1) = 1
              f#!_A(x1) = x1 + 2
              g!_A(x1) = 1
              f!_A(x1) = 0
              h!_A(x1) = 2
        
  
    partial status:
  
      pi(f#) = []
      pi(g) = []
      pi(f) = [1]
      pi(h) = []

The following rules are removed:

  p1, p2

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            |f'|#_A(x1,x2,x3) = 0
            s_A(x1) = x1 + 7
            |f'|#!_A(x1,x2,x3) = max{1, x1 - 4, x2 - 3, x3 - 8}
            s!_A(x1) = max{4, x1 - 3}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              |f'|#_A(x1,x2,x3) = 0
              s_A(x1) = x1 + 1
              |f'|#!_A(x1,x2,x3) = 1
              s!_A(x1) = 0
        
  
    partial status:
  
      pi(|f'|#) = [1, 2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.