NO

Problem:
 nats() -> cons(0(),n__incr(n__nats()))
 pairs() -> cons(0(),n__incr(n__odds()))
 odds() -> incr(pairs())
 incr(cons(X,XS)) -> cons(s(X),n__incr(activate(XS)))
 head(cons(X,XS)) -> X
 tail(cons(X,XS)) -> activate(XS)
 incr(X) -> n__incr(X)
 nats() -> n__nats()
 odds() -> n__odds()
 activate(n__incr(X)) -> incr(activate(X))
 activate(n__nats()) -> nats()
 activate(n__odds()) -> odds()
 activate(X) -> X

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [cons](x0, x1) = x0 + x1,
   
   [incr](x0) = 4x0,
   
   [0] = 0,
   
   [n__incr](x0) = 4x0,
   
   [pairs] = 0,
   
   [activate](x0) = x0,
   
   [nats] = 0,
   
   [odds] = 0,
   
   [tail](x0) = 6x0 + 1,
   
   [head](x0) = x0 + 1,
   
   [s](x0) = 4x0,
   
   [n__odds] = 0,
   
   [n__nats] = 0
  orientation:
   nats() = 0 >= 0 = cons(0(),n__incr(n__nats()))
   
   pairs() = 0 >= 0 = cons(0(),n__incr(n__odds()))
   
   odds() = 0 >= 0 = incr(pairs())
   
   incr(cons(X,XS)) = 4X + 4XS >= 4X + 4XS = cons(s(X),n__incr(activate(XS)))
   
   head(cons(X,XS)) = X + XS + 1 >= X = X
   
   tail(cons(X,XS)) = 6X + 6XS + 1 >= XS = activate(XS)
   
   incr(X) = 4X >= 4X = n__incr(X)
   
   nats() = 0 >= 0 = n__nats()
   
   odds() = 0 >= 0 = n__odds()
   
   activate(n__incr(X)) = 4X >= 4X = incr(activate(X))
   
   activate(n__nats()) = 0 >= 0 = nats()
   
   activate(n__odds()) = 0 >= 0 = odds()
   
   activate(X) = X >= X = X
  problem:
   nats() -> cons(0(),n__incr(n__nats()))
   pairs() -> cons(0(),n__incr(n__odds()))
   odds() -> incr(pairs())
   incr(cons(X,XS)) -> cons(s(X),n__incr(activate(XS)))
   incr(X) -> n__incr(X)
   nats() -> n__nats()
   odds() -> n__odds()
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats()) -> nats()
   activate(n__odds()) -> odds()
   activate(X) -> X
  Unfolding Processor:
   loop length: 4
   terms:
    incr(cons(X,n__incr(n__nats())))
    cons(s(X),n__incr(activate(n__incr(n__nats()))))
    cons(s(X),n__incr(incr(activate(n__nats()))))
    cons(s(X),n__incr(incr(nats())))
   context: cons(s(X),n__incr([]))
   substitution:
    X -> 0()
   Qed