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:

  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:
          
            cond#_A(x1,x2) = max{0, x1 - 25, x2 - 2}
            true_A = 9
            odd_A(x1) = max{3, x1 + 1}
            p_A(x1) = max{1, x1 - 7}
            |0|_A = 0
            false_A = 3
            s_A(x1) = max{18, x1 + 7}
            cond#!_A(x1,x2) = max{x1, x2 + 5}
            true!_A = 11
            odd!_A(x1) = 10
            p!_A(x1) = max{21, x1 + 19}
            |0|!_A = 20
            false!_A = 1
            s!_A(x1) = 12
        
        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() = 2
              odd_A(x1) = x1 + 1
              p_A(x1) = x1 + 1
              |0|_A() = 1
              false_A() = 0
              s_A(x1) = x1 + 1
              cond#!_A(x1,x2) = x1 + 3
              true!_A() = 2
              odd!_A(x1) = 1
              p!_A(x1) = 4
              |0|!_A() = 3
              false!_A() = 0
              s!_A(x1) = 3
        
  
    partial status:
  
      pi(cond#) = []
      pi(true) = []
      pi(odd) = [1]
      pi(p) = [1]
      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(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 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:
          
            odd#_A(x1) = 0
            s_A(x1) = max{0, x1 - 1}
            odd#!_A(x1) = 0
            s!_A(x1) = max{0, x1 - 13}
        
        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
              odd#!_A(x1) = 3
              s!_A(x1) = 3
        
  
    partial status:
  
      pi(odd#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.