YES

We show the termination of the TRS R:

  cond(true(),x) -> cond(odd(x),p(x))
  odd(|0|()) -> false()
  odd(s(|0|())) -> true()
  odd(s(s(x))) -> odd(x)
  p(|0|()) -> |0|()
  p(s(x)) -> x

-- SCC decomposition.

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

p1: cond#(true(),x) -> cond#(odd(x),p(x))
p2: cond#(true(),x) -> odd#(x)
p3: cond#(true(),x) -> p#(x)
p4: odd#(s(s(x))) -> odd#(x)

and R consists of:

r1: cond(true(),x) -> cond(odd(x),p(x))
r2: odd(|0|()) -> false()
r3: odd(s(|0|())) -> true()
r4: odd(s(s(x))) -> odd(x)
r5: p(|0|()) -> |0|()
r6: p(s(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1}
  {p4}


-- Reduction pair.

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

p1: cond#(true(),x) -> cond#(odd(x),p(x))

and R consists of:

r1: cond(true(),x) -> cond(odd(x),p(x))
r2: odd(|0|()) -> false()
r3: odd(s(|0|())) -> true()
r4: odd(s(s(x))) -> odd(x)
r5: p(|0|()) -> |0|()
r6: p(s(x)) -> x

The set of usable rules consists of

  r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cond#_A(x1,x2) = max{38, x1, x2 + 18}
          true_A = 39
          odd_A(x1) = max{32, x1 + 17}
          p_A(x1) = max{19, x1 - 7}
          |0|_A = 18
          false_A = 33
          s_A(x1) = max{23, x1 + 8}
    
      precedence:
      
        cond# = true = odd = p = |0| = false = s
      
      partial status:
    
        pi(cond#) = [1, 2]
        pi(true) = []
        pi(odd) = [1]
        pi(p) = []
        pi(|0|) = []
        pi(false) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cond#_A(x1,x2) = max{2, x1 - 5, x2}
          true_A = 7
          odd_A(x1) = max{10, x1 + 6}
          p_A(x1) = 3
          |0|_A = 6
          false_A = 11
          s_A(x1) = x1 + 2
    
      precedence:
      
        cond# = true = odd = p = |0| = false = s
      
      partial status:
    
        pi(cond#) = [2]
        pi(true) = []
        pi(odd) = [1]
        pi(p) = []
        pi(|0|) = []
        pi(false) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: odd#(s(s(x))) -> odd#(x)

and R consists of:

r1: cond(true(),x) -> cond(odd(x),p(x))
r2: odd(|0|()) -> false()
r3: odd(s(|0|())) -> true()
r4: odd(s(s(x))) -> odd(x)
r5: p(|0|()) -> |0|()
r6: p(s(x)) -> x

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          odd#_A(x1) = max{2, x1 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        odd# = s
      
      partial status:
    
        pi(odd#) = [1]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          odd#_A(x1) = x1 + 1
          s_A(x1) = x1
    
      precedence:
      
        odd# = s
      
      partial status:
    
        pi(odd#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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