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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          evenodd#_A(x1,x2) = x1 + (0,6)
          |0|_A() = (1,5)
          s_A(x1) = ((1,0),(1,1)) x1 + (0,7)
          not_A(x1) = (3,3)
          true_A() = (1,2)
          false_A() = (2,1)
          evenodd_A(x1,x2) = ((1,0),(0,0)) x2 + (3,4)
          rand_A(x1) = ((1,0),(0,0)) x1 + (1,8)
  
    precedence:
    
      |0| = not = true = false = evenodd = rand > evenodd# = s
    
    partial status:
  
      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)