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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        cond#_A(x1,x2) = max{17, x1 + 12}
        true_A = 1
        odd_A(x1) = 5
        p_A(x1) = max{3, x1 - 5}
        |0|_A = 2
        false_A = 1
        s_A(x1) = max{13, x1 + 6}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          cond#_A(x1,x2) = max{0, x1 - 10, x2 - 5}
          true_A = 8
          odd_A(x1) = max{0, x1 - 1}
          p_A(x1) = max{4, x1 - 3}
          |0|_A = 0
          false_A = 0
          s_A(x1) = max{9, x1 + 3}
          cond#!_A(x1,x2) = max{x1 - 1, x2 + 2}
          true!_A = 5
          odd!_A(x1) = max{4, x1 + 1}
          p!_A(x1) = max{8, x1 - 6}
          |0|!_A = 6
          false!_A = 2
          s!_A(x1) = max{1, x1 - 10}
  
    partial status:
  
      pi(cond#) = [1]
      pi(true) = []
      pi(odd) = []
      pi(p) = []
      pi(|0|) = []
      pi(false) = []
      pi(s) = [1]

The following rules are removed:

  p1

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

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        odd#_A(x1) = max{13, x1 - 15}
        s_A(x1) = max{14, x1 + 6}
        cond_A(x1,x2) = max{14, x1 - 14, x2 - 4}
        true_A = 1
        odd_A(x1) = max{13, x1 - 15}
        p_A(x1) = max{3, x1 - 5}
        |0|_A = 2
        false_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          odd#_A(x1) = x1
          s_A(x1) = max{12, x1 + 8}
          cond_A(x1,x2) = max{0, x1 - 3, x2 - 14}
          true_A = 1
          odd_A(x1) = max{2, x1 - 13}
          p_A(x1) = max{13, x1 - 1}
          |0|_A = 0
          false_A = 0
          odd#!_A(x1) = max{1, x1 - 17}
          s!_A(x1) = max{3, x1 - 13}
          cond!_A(x1,x2) = max{4, x1 - 3, x2 - 16}
          true!_A = 1
          odd!_A(x1) = max{2, x1 - 17}
          p!_A(x1) = max{3, x1 - 17}
          |0|!_A = 2
          false!_A = 0
  
    partial status:
  
      pi(odd#) = []
      pi(s) = [1]
      pi(cond) = []
      pi(true) = []
      pi(odd) = []
      pi(p) = []
      pi(|0|) = []
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.