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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 2
          g_A(x1) = x1
          f_A(x1) = x1
          h_A(x1) = x1 + 2
          |f'|_A(x1,x2,x3) = 1
          s_A(x1) = 0
          rand_A(x1) = x1
    
      precedence:
      
        f# = g = f = h = |f'| = s = rand
      
      partial status:
    
        pi(f#) = []
        pi(g) = []
        pi(f) = []
        pi(h) = []
        pi(|f'|) = []
        pi(s) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{5, x1 + 4}
          g_A(x1) = x1 + 3
          f_A(x1) = x1
          h_A(x1) = 3
          |f'|_A(x1,x2,x3) = 1
          s_A(x1) = 0
          rand_A(x1) = x1
    
      precedence:
      
        f# = g = f = h = |f'| = s = rand
      
      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.

-- SCC decomposition.

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

p1: 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 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#(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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{4, x1 + 1}
          g_A(x1) = max{5, x1 + 2}
          f_A(x1) = max{3, x1}
          h_A(x1) = 3
          |f'|_A(x1,x2,x3) = max{x1 + 4, x3 + 5}
          s_A(x1) = 0
          rand_A(x1) = x1 + 1
    
      precedence:
      
        f > f# = g = h = |f'| = s = rand
      
      partial status:
    
        pi(f#) = [1]
        pi(g) = [1]
        pi(f) = [1]
        pi(h) = []
        pi(|f'|) = []
        pi(s) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{1, x1}
          g_A(x1) = max{6, x1 + 2}
          f_A(x1) = max{3, x1}
          h_A(x1) = 6
          |f'|_A(x1,x2,x3) = 1
          s_A(x1) = 2
          rand_A(x1) = 0
    
      precedence:
      
        f# = g = f = h = |f'| = s = rand
      
      partial status:
    
        pi(f#) = [1]
        pi(g) = [1]
        pi(f) = [1]
        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:

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