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-minimaldependency 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-minimaldependency 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,1,1)) x1
      c_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,1,0)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,0,0)) x2 + (0,1,1,1)
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x1 + (0,0,1,1)
      g#_A(x1) = x1 + (0,0,1,3)
      f_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1
      g_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1
      rand_A(x1) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,1,1,0)

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.

-- SCC decomposition.

Consider the non-minimaldependency 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the non-minimaldependency 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      g#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,1,1)) x1
      c_A(x1,x2) = x1 + ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,1,0,0)) x2 + (1,1,1,0)
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,1,1,1)) x1 + (0,1,1,1)
      f_A(x1) = x1 + (0,1,0,0)
      g_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(0,0,0,0)) x1 + (0,0,0,1)
      rand_A(x1) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,1,1,0)

The next rules are strictly ordered:

  p1

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