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, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1) = x1
          g_A(x1) = x1 + 2
          f_A(x1) = x1
          h_A(x1) = 1
          |f'|_A(x1,x2,x3) = 2
          s_A(x1) = 1
  
    base order 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) = 3
            f_A(x1) = 3
            h_A(x1) = 1
            |f'|_A(x1,x2,x3) = x1 + x3
            s_A(x1) = x1 + 1
            f#!_A(x1) = x1 + 2
            g!_A(x1) = 3
            f!_A(x1) = 4
            h!_A(x1) = x1
            |f'|!_A(x1,x2,x3) = x1 + x3
            s!_A(x1) = 0
  
    partial status:
  
      pi(f#) = [1]
      pi(g) = []
      pi(f) = []
      pi(h) = []
      pi(|f'|) = []
      pi(s) = []

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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          |f'|#_A(x1,x2,x3) = 1
          s_A(x1) = x1 + 2
  
    base order 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) = x1 + x2
            s!_A(x1) = 2
  
    partial status:
  
      pi(|f'|#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.