YES

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

  R:
  not(true()) -> false()
  not(false()) -> true()
  evenodd(x,|0|()) -> not(evenodd(x,s(|0|())))
  evenodd(|0|(),s(|0|())) -> false()
  evenodd(s(x),s(|0|())) -> evenodd(x,|0|())

  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: evenodd#(x,|0|()) -> not#(evenodd(x,s(|0|())))
p2: evenodd#(x,|0|()) -> evenodd#(x,s(|0|()))
p3: evenodd#(s(x),s(|0|())) -> evenodd#(x,|0|())

and R consists of:

r1: not(true()) -> false()
r2: not(false()) -> true()
r3: evenodd(x,|0|()) -> not(evenodd(x,s(|0|())))
r4: evenodd(|0|(),s(|0|())) -> false()
r5: evenodd(s(x),s(|0|())) -> evenodd(x,|0|())
r6: rand(x) -> x
r7: 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: evenodd#(x,|0|()) -> evenodd#(x,s(|0|()))
p2: evenodd#(s(x),s(|0|())) -> evenodd#(x,|0|())

and R consists of:

r1: not(true()) -> false()
r2: not(false()) -> true()
r3: evenodd(x,|0|()) -> not(evenodd(x,s(|0|())))
r4: evenodd(|0|(),s(|0|())) -> false()
r5: evenodd(s(x),s(|0|())) -> evenodd(x,|0|())
r6: rand(x) -> x
r7: rand(x) -> rand(s(x))

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        evenodd#_A(x1,x2) = x1 + ((1,1),(0,0)) x2
        |0|_A() = (1,1)
        s_A(x1) = ((1,1),(0,0)) x1
        not_A(x1) = ((0,1),(0,0)) x1 + (1,1)
        true_A() = (1,1)
        false_A() = (1,1)
        evenodd_A(x1,x2) = ((1,0),(1,0)) x2 + (3,0)
        rand_A(x1) = ((1,1),(0,1)) x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        evenodd#_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,1),(0,0)) x2
        |0|_A() = (1,2)
        s_A(x1) = ((1,1),(0,0)) x1 + (1,1)
        not_A(x1) = (2,3)
        true_A() = (1,4)
        false_A() = (5,4)
        evenodd_A(x1,x2) = x2
        rand_A(x1) = (0,0)
    

The next rules are strictly ordered:

  p1, p2

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