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. lexicographic path order with precedence:
    
      precedence:
      
        rand > |0| > not > evenodd > true > evenodd# > s > false
      
      argument filter:
    
        pi(evenodd#) = [1, 2]
        pi(|0|) = []
        pi(s) = 1
        pi(not) = []
        pi(true) = []
        pi(false) = []
        pi(evenodd) = 2
        pi(rand) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        rand > not > true > |0| > evenodd# > s > false > evenodd
      
      argument filter:
    
        pi(evenodd#) = [1]
        pi(|0|) = []
        pi(s) = [1]
        pi(not) = []
        pi(true) = []
        pi(false) = []
        pi(evenodd) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: evenodd#(x,|0|()) -> evenodd#(x,s(|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:

  (no SCCs)