YES

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

  R:
  f(|0|()) -> true()
  f(|1|()) -> false()
  f(s(x)) -> f(x)
  if(true(),s(x),s(y)) -> s(x)
  if(false(),s(x),s(y)) -> s(y)
  g(x,c(y)) -> c(g(x,y))
  g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),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#(s(x)) -> f#(x)
p2: g#(x,c(y)) -> g#(x,y)
p3: g#(x,c(y)) -> g#(x,if(f(x),c(g(s(x),y)),c(y)))
p4: g#(x,c(y)) -> if#(f(x),c(g(s(x),y)),c(y))
p5: g#(x,c(y)) -> f#(x)
p6: g#(x,c(y)) -> g#(s(x),y)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2, p6}
  {p1}


-- Reduction pair.

Consider the non-minimal dependency pair problem (P, R), where P consists of

p1: g#(x,c(y)) -> g#(s(x),y)
p2: g#(x,c(y)) -> g#(x,y)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      g#_A(x1,x2) = x2
      c_A(x1) = max{10, x1 + 5}
      s_A(x1) = 0
      f_A(x1) = 2
      |0|_A = 0
      true_A = 1
      |1|_A = 0
      false_A = 1
      if_A(x1,x2,x3) = max{2, x1, x2 - 6, x3 - 8}
      g_A(x1,x2) = max{3, x2}
      rand_A(x1) = x1 + 1
    
    2. max/plus interpretations on natural numbers:
    
      g#_A(x1,x2) = 0
      c_A(x1) = 0
      s_A(x1) = 1
      f_A(x1) = 1
      |0|_A = 2
      true_A = 0
      |1|_A = 0
      false_A = 0
      if_A(x1,x2,x3) = 2
      g_A(x1,x2) = 1
      rand_A(x1) = 0
    

The next rules are strictly ordered:

  p1, p2

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)) -> f#(x)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      f#_A(x1) = x1
      s_A(x1) = x1
      f_A(x1) = 1
      |0|_A = 1
      true_A = 0
      |1|_A = 1
      false_A = 0
      if_A(x1,x2,x3) = max{x1 - 4, x2 + 1, x3 + 2}
      g_A(x1,x2) = x1 + 4
      c_A(x1) = max{0, x1 - 1}
      rand_A(x1) = x1 + 1
    
    2. max/plus interpretations on natural numbers:
    
      f#_A(x1) = x1
      s_A(x1) = x1 + 1
      f_A(x1) = 1
      |0|_A = 0
      true_A = 2
      |1|_A = 0
      false_A = 0
      if_A(x1,x2,x3) = 2
      g_A(x1,x2) = 1
      c_A(x1) = 0
      rand_A(x1) = 0
    

The next rules are strictly ordered:

  p1

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