YES

We show the termination of the relative TRS R/S:

  R:
  f(g(x)) -> g(f(f(x)))
  f(h(x)) -> h(g(x))
  |f'|(s(x),y,y) -> |f'|(y,x,s(x))

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

Consider the non-minimal 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))
r4: rand(x) -> x
r5: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p1, p2}
  {p3}


-- Reduction pair.

Consider the non-minimal 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))
r4: rand(x) -> x
r5: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1) = ((1,0),(1,0)) x1 + (1,0)
          g_A(x1) = ((1,0),(1,0)) x1 + (4,2)
          f_A(x1) = ((1,0),(1,0)) x1 + (0,3)
          h_A(x1) = (1,1)
          |f'|_A(x1,x2,x3) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x3 + (1,1)
          s_A(x1) = x1
          rand_A(x1) = ((1,0),(1,1)) x1 + (1,1)
  
    precedence:
    
      f# = g = f = h = rand > |f'| > s
    
    partial status:
  
      pi(f#) = []
      pi(g) = []
      pi(f) = []
      pi(h) = []
      pi(|f'|) = []
      pi(s) = []
      pi(rand) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the non-minimal 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))
r4: rand(x) -> x
r5: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the non-minimal 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))
r4: rand(x) -> x
r5: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          g_A(x1) = ((1,0),(0,0)) x1 + (3,3)
          f_A(x1) = ((1,0),(1,0)) x1 + (0,2)
          h_A(x1) = (2,1)
          |f'|_A(x1,x2,x3) = x1 + x3 + (1,1)
          s_A(x1) = ((0,0),(1,0)) x1
          rand_A(x1) = ((1,0),(1,1)) x1 + (1,1)
  
    precedence:
    
      g = f > |f'| = s = rand > f# = h
    
    partial status:
  
      pi(f#) = []
      pi(g) = []
      pi(f) = []
      pi(h) = []
      pi(|f'|) = []
      pi(s) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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))
r4: rand(x) -> x
r5: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          |f'|#_A(x1,x2,x3) = x1 + x2 + (1,2)
          s_A(x1) = x1 + (0,1)
          f_A(x1) = (3,6)
          g_A(x1) = ((0,0),(1,0)) x1 + (1,2)
          h_A(x1) = (2,1)
          |f'|_A(x1,x2,x3) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + x3 + (0,1)
          rand_A(x1) = ((1,0),(1,0)) x1 + (1,2)
  
    precedence:
    
      h = rand > f = g > |f'|# = s = |f'|
    
    partial status:
  
      pi(|f'|#) = [1, 2]
      pi(s) = []
      pi(f) = []
      pi(g) = []
      pi(h) = []
      pi(|f'|) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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