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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = ((0,0),(1,0)) x2
      c_A(x1) = ((1,0),(1,1)) x1 + (2,1)
      f_A(x1,x2) = ((1,0),(1,1)) x1 + (1,1)
      s_A(x1) = ((1,0),(1,1)) x1 + (0,1)
      rand_A(x1) = x1 + (1,0)

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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = x1
      s_A(x1) = ((1,0),(1,1)) x1 + (0,1)
      c_A(x1) = x1 + (2,1)
      f_A(x1,x2) = x1 + (1,0)
      rand_A(x1) = x1 + (1,0)

The next rules are strictly ordered:

  p1

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