YES

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

  R:
  g(x,y) -> x
  g(x,y) -> y
  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: g(x,y) -> x
r2: g(x,y) -> y
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#(s(x),y,y) -> f#(y,x,s(x))

and R consists of:

r1: g(x,y) -> x
r2: g(x,y) -> y
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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2,x3) = x1 + x2
        s_A(x1) = x1
        g_A(x1,x2) = x1 + x2 + (1,0)
        f_A(x1,x2,x3) = x1 + x3
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2,x3) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2
        s_A(x1) = x1 + (1,1)
        g_A(x1,x2) = x1 + x2 + (1,0)
        f_A(x1,x2,x3) = ((1,1),(0,1)) x1 + ((1,1),(0,1)) x3
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2,x3) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2
        s_A(x1) = ((1,1),(1,0)) x1 + (1,1)
        g_A(x1,x2) = (0,0)
        f_A(x1,x2,x3) = x1 + x3
        rand_A(x1) = (0,0)
    

The next rules are strictly ordered:

  p1

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