YES

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

  R:
  f(c(s(x),y)) -> f(c(x,s(y)))
  f(c(s(x),s(y))) -> g(c(x,y))
  g(c(x,s(y))) -> g(c(s(x),y))
  g(c(s(x),s(y))) -> f(c(x,y))

  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#(c(s(x),y)) -> f#(c(x,s(y)))
p2: f#(c(s(x),s(y))) -> g#(c(x,y))
p3: g#(c(x,s(y))) -> g#(c(s(x),y))
p4: g#(c(s(x),s(y))) -> f#(c(x,y))

and R consists of:

r1: f(c(s(x),y)) -> f(c(x,s(y)))
r2: f(c(s(x),s(y))) -> g(c(x,y))
r3: g(c(x,s(y))) -> g(c(s(x),y))
r4: g(c(s(x),s(y))) -> f(c(x,y))
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: f#(c(s(x),y)) -> f#(c(x,s(y)))
p2: f#(c(s(x),s(y))) -> g#(c(x,y))
p3: g#(c(s(x),s(y))) -> f#(c(x,y))
p4: g#(c(x,s(y))) -> g#(c(s(x),y))

and R consists of:

r1: f(c(s(x),y)) -> f(c(x,s(y)))
r2: f(c(s(x),s(y))) -> g(c(x,y))
r3: g(c(x,s(y))) -> g(c(s(x),y))
r4: g(c(s(x),s(y))) -> f(c(x,y))
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1) = ((1,0),(0,0)) x1 + (1,2)
            c_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (0,1)
            s_A(x1) = x1
            g#_A(x1) = ((1,0),(0,0)) x1 + (1,2)
            f_A(x1) = ((0,0),(1,1)) x1 + (1,1)
            g_A(x1) = ((0,0),(1,0)) x1 + (1,2)
            rand_A(x1) = ((1,1),(0,1)) x1 + (1,0)
    
      precedence:
      
        f# = c = s = g# = f = g = rand
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        pi(s) = []
        pi(g#) = []
        pi(f) = []
        pi(g) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1) = ((1,1),(0,1)) x1 + (1,0)
            c_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (2,2)
            s_A(x1) = ((1,1),(0,0)) x1 + (1,1)
            g#_A(x1) = ((1,0),(0,0)) x1 + (0,2)
            f_A(x1) = (4,2)
            g_A(x1) = (4,2)
            rand_A(x1) = (0,0)
    
      precedence:
      
        c = g# = rand > f# > s = f = g
      
      partial status:
    
        pi(f#) = [1]
        pi(c) = []
        pi(s) = []
        pi(g#) = []
        pi(f) = []
        pi(g) = []
        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#(c(s(x),y)) -> f#(c(x,s(y)))
p2: g#(c(s(x),s(y))) -> f#(c(x,y))
p3: g#(c(x,s(y))) -> g#(c(s(x),y))

and R consists of:

r1: f(c(s(x),y)) -> f(c(x,s(y)))
r2: f(c(s(x),s(y))) -> g(c(x,y))
r3: g(c(x,s(y))) -> g(c(s(x),y))
r4: g(c(s(x),s(y))) -> f(c(x,y))
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p3}
  {p1}


-- Reduction pair.

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

p1: g#(c(x,s(y))) -> g#(c(s(x),y))

and R consists of:

r1: f(c(s(x),y)) -> f(c(x,s(y)))
r2: f(c(s(x),s(y))) -> g(c(x,y))
r3: g(c(x,s(y))) -> g(c(s(x),y))
r4: g(c(s(x),s(y))) -> f(c(x,y))
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{8, x1 + 2}
          c_A(x1,x2) = x2 + 7
          s_A(x1) = x1
          f_A(x1) = 1
          g_A(x1) = 1
          rand_A(x1) = x1 + 1
    
      precedence:
      
        g# = c = s = f = g = rand
      
      partial status:
    
        pi(g#) = [1]
        pi(c) = [2]
        pi(s) = [1]
        pi(f) = []
        pi(g) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{0, x1 - 16}
          c_A(x1,x2) = max{17, x2 + 8}
          s_A(x1) = x1 + 14
          f_A(x1) = 43
          g_A(x1) = 43
          rand_A(x1) = 9
    
      precedence:
      
        c > g# > rand > s > f = g
      
      partial status:
    
        pi(g#) = []
        pi(c) = [2]
        pi(s) = []
        pi(f) = []
        pi(g) = []
        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#(c(s(x),y)) -> f#(c(x,s(y)))

and R consists of:

r1: f(c(s(x),y)) -> f(c(x,s(y)))
r2: f(c(s(x),s(y))) -> g(c(x,y))
r3: g(c(x,s(y))) -> g(c(s(x),y))
r4: g(c(s(x),s(y))) -> f(c(x,y))
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

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{1, x1}
          c_A(x1,x2) = max{x1 + 10, x2 + 5}
          s_A(x1) = max{4, x1}
          f_A(x1) = max{3, x1 - 13}
          g_A(x1) = max{3, x1 - 13}
          rand_A(x1) = max{6, x1 + 1}
    
      precedence:
      
        c > f# = s > f = g = rand
      
      partial status:
    
        pi(f#) = [1]
        pi(c) = [1, 2]
        pi(s) = [1]
        pi(f) = []
        pi(g) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = 2
          c_A(x1,x2) = max{x1 + 6, x2 + 6}
          s_A(x1) = x1
          f_A(x1) = 6
          g_A(x1) = 6
          rand_A(x1) = 1
    
      precedence:
      
        f# = c > s > f = g = rand
      
      partial status:
    
        pi(f#) = []
        pi(c) = [1]
        pi(s) = [1]
        pi(f) = []
        pi(g) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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