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:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1) = x1 + 3
        g_A(x1) = x1 + 4
        f_A(x1) = x1
        h_A(x1) = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1) = max{0, x1 - 4}
          g_A(x1) = max{2, x1}
          f_A(x1) = max{3, x1}
          h_A(x1) = max{4, x1 - 3}
          f#!_A(x1) = 1
          g!_A(x1) = max{2, x1 - 12}
          f!_A(x1) = max{3, x1 - 5}
          h!_A(x1) = max{1, x1 - 11}
  
    partial status:
  
      pi(f#) = [1]
      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:
  
      max/plus interpretation on natural numbers:
      
        |f'|#_A(x1,x2,x3) = max{x1 + 9, x2 + 10}
        s_A(x1) = max{8, x1 + 2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          |f'|#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 2, x3 - 2}
          s_A(x1) = max{1, x1}
          |f'|#!_A(x1,x2,x3) = max{2, x1 - 4, x2 - 1, x3 - 3}
          s!_A(x1) = max{1, x1 - 5}
  
    partial status:
  
      pi(|f'|#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.