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:

  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) = ((1,0),(1,1)) x1 + (0,1)
          f_A(x1,x2,x3) = x1 + x3 + (1,1)
          rand_A(x1) = ((1,0),(1,0)) x1 + (1,2)
  
    precedence:
    
      f > rand > f# = s
    
    partial status:
  
      pi(f#) = []
      pi(s) = []
      pi(f) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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