YES
We show the termination of the TRS 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|())


-- SCC decomposition.

Consider the 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|())

The estimated dependency graph contains the following SCCs:

  {p2, p3}


-- Reduction pair.

Consider the 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|())

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        evenodd#_A(x1,x2) = max{0, x2 - 11}
        |0|_A = 2
        s_A(x1) = max{3, x1 - 10}
        not_A(x1) = max{3, x1 - 5}
        true_A = 0
        false_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          evenodd#_A(x1,x2) = max{0, x1 - 20, x2 - 16}
          |0|_A = 12
          s_A(x1) = max{15, x1 + 3}
          not_A(x1) = max{1, x1 - 2}
          true_A = 0
          false_A = 1
          evenodd#!_A(x1,x2) = max{x1 + 3, x2 - 13}
          |0|!_A = 4
          s!_A(x1) = max{5, x1 - 5}
          not!_A(x1) = max{2, x1 - 2}
          true!_A = 0
          false!_A = 1
  
    partial status:
  
      pi(evenodd#) = []
      pi(|0|) = []
      pi(s) = []
      pi(not) = []
      pi(true) = []
      pi(false) = []

The following rules are removed:

  p2



-- SCC decomposition.

Consider the 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|())

The estimated dependency graph contains the following SCCs:

  (no SCCs)