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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            evenodd#_A(x1,x2) = max{0, x1 - 18, x2 - 11}
            |0|_A = 2
            s_A(x1) = x1 + 8
            evenodd#!_A(x1,x2) = max{5, x1 - 16, x2 - 11}
            |0|!_A = 3
            s!_A(x1) = max{4, x1 - 16}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            evenodd#_A(x1,x2) = max{0, x1 - 12, x2 - 7}
            |0|_A = 2
            s_A(x1) = max{6, x1 + 4}
            evenodd#!_A(x1,x2) = max{3, x1 - 16, x2 - 16}
            |0|!_A = 1
            s!_A(x1) = max{2, x1 - 15}
        
  
    partial status:
  
      pi(evenodd#) = [1]
      pi(|0|) = []
      pi(s) = [1]

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)