YES

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        f#_A(x1,x2,x3) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,1,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,1,1,0),(1,1,0,1)) x2 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)) x3
        s_A(x1) = ((1,0,0,0),(1,1,0,0),(0,1,0,0),(1,1,1,0)) x1 + (0,1,2,0)
        f_A(x1,x2,x3) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,0,0)) x2 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x3
        rand_A(x1) = x1 + (1,0,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        rand > f > s > f#
      
      argument filter:
    
        pi(f#) = 2
        pi(s) = []
        pi(f) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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