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{8, x1 - 2, x2 + 3}
        true_A = 12
        odd_A(x1) = max{11, x1 + 2}
        p_A(x1) = max{4, x1 - 2}
        |0|_A = 3
        false_A = 2
        s_A(x1) = max{11, x1 + 3}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            cond#_A(x1,x2) = 0
            true_A() = 1
            odd_A(x1) = 1
            p_A(x1) = x1 + 1
            |0|_A() = 1
            false_A() = 0
            s_A(x1) = x1 + 1
            cond#!_A(x1,x2) = 2
            true!_A() = 3
            odd!_A(x1) = 1
            p!_A(x1) = 1
            |0|!_A() = 2
            false!_A() = 0
            s!_A(x1) = 2
  
    partial status:
  
      pi(cond#) = []
      pi(true) = []
      pi(odd) = [1]
      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

  r5

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        odd#_A(x1) = 0
        s_A(x1) = max{3, x1 - 4}
        p_A(x1) = max{7, x1 - 9}
        |0|_A = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            odd#_A(x1) = x1
            s_A(x1) = x1 + 1
            p_A(x1) = x1 + 4
            |0|_A() = 1
            odd#!_A(x1) = x1 + 4
            s!_A(x1) = 3
            p!_A(x1) = 3
            |0|!_A() = 1
  
    partial status:
  
      pi(odd#) = []
      pi(s) = []
      pi(p) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.