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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1) = max{7, x1 + 3}
        g_A(x1) = max{8, x1 + 5}
        f_A(x1) = max{3, x1}
        h_A(x1) = 4
        |f'|_A(x1,x2,x3) = max{0, x1 - 1, x3 - 1}
        s_A(x1) = max{2, x1 - 1}
  
    base order 2:
  
      precedence:
      
        g = |f'| = s > f# = f > h
      
  
    partial status:
  
      pi(f#) = [1]
      pi(g) = []
      pi(f) = [1]
      pi(h) = []
      pi(|f'|) = []
      pi(s) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: f#(g(x)) -> f#(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 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#(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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1) = max{4, x1 + 3}
        g_A(x1) = x1 + 5
        f_A(x1) = x1
        h_A(x1) = 1
        |f'|_A(x1,x2,x3) = max{3, x1, x2, x3 - 1}
        s_A(x1) = max{2, x1}
  
    base order 2:
  
      precedence:
      
        g = |f'| = s > f# > f > h
      
  
    partial status:
  
      pi(f#) = [1]
      pi(g) = []
      pi(f) = [1]
      pi(h) = []
      pi(|f'|) = []
      pi(s) = []

The following rules are removed:

  p1

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

  r3

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        |f'|#_A(x1,x2,x3) = max{3, x1, x2 + 1, x3 - 5}
        s_A(x1) = x1 + 4
        |f'|_A(x1,x2,x3) = max{5, x1 - 5, x2 - 5, x3 - 8}
  
    base order 2:
  
      precedence:
      
        |f'| > |f'|# = s
      
  
    partial status:
  
      pi(|f'|#) = [1]
      pi(s) = [1]
      pi(|f'|) = []

The following rules are removed:

  p1

Then no dependency pair remains.