YES

We show the termination of the TRS R:

  cond(true(),x) -> cond(odd(x),p(p(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(p(p(x))))
p2: cond#(true(),x) -> odd#(x)
p3: cond#(true(),x) -> p#(p(p(x)))
p4: cond#(true(),x) -> p#(p(x))
p5: cond#(true(),x) -> p#(x)
p6: odd#(s(s(x))) -> odd#(x)

and R consists of:

r1: cond(true(),x) -> cond(odd(x),p(p(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}
  {p6}


-- Reduction pair.

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

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

and R consists of:

r1: cond(true(),x) -> cond(odd(x),p(p(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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cond#_A(x1,x2) = max{x1 + 3, x2 + 19}
        true_A = 21
        odd_A(x1) = x1 + 15
        p_A(x1) = max{4, x1 - 6}
        |0|_A = 3
        false_A = 2
        s_A(x1) = max{22, x1 + 7}
  
    precedence:
    
      cond# = true = odd = p = |0| = false = s
    
    partial status:
  
      pi(cond#) = [1]
      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(p(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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        odd#_A(x1) = x1 + 5
        s_A(x1) = x1 + 2
  
    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.