YES

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3}


-- Reduction pair.

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

p1: f#(x,c(y)) -> f#(y,y)

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 8, x2 + 5}
          c_A(x1) = x1 + 4
          f_A(x1,x2) = x2 + 3
          s_A(x1) = max{0, x1 - 4}
          rand_A(x1) = x1
    
      precedence:
      
        f# = c = f = s = rand
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        pi(f) = []
        pi(s) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = 8
          c_A(x1) = x1 + 7
          f_A(x1,x2) = 1
          s_A(x1) = 0
          rand_A(x1) = x1 + 6
    
      precedence:
      
        rand > s > f# = c = f
      
      partial status:
    
        pi(f#) = []
        pi(c) = [1]
        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) -> f#(x,s(c(y)))

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 2, x2 - 1}
          s_A(x1) = x1
          c_A(x1) = max{3, x1}
          f_A(x1,x2) = 1
          rand_A(x1) = x1 + 2
    
      precedence:
      
        f# > c = f > s = rand
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(c) = [1]
        pi(f) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{11, x1 + 1}
          s_A(x1) = x1 + 10
          c_A(x1) = max{13, x1 + 7}
          f_A(x1,x2) = 3
          rand_A(x1) = 3
    
      precedence:
      
        c > s = f > f# = rand
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = []
        pi(c) = []
        pi(f) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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