NO

Problem:
 sq(0(x)) -> p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x)))))))))))))))))
 sq(s(x)) ->
 s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x)))))))))))))))))))))))))))))))))
 twice(0(x)) -> p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x)))))))))))))))))))))))))))
 twice(s(x)) -> p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x)))))))))))))))
 p(p(s(x))) -> p(x)
 p(s(x)) -> x
 p(0(x)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x))))))))))))
 0(x) -> x

Proof:
 Nonconfluence Processor:
  terms: p(f5()) *<- p(0(f5())) ->* 0(s(s(s(s(s(s(s(s(s(s(s(f5()))))))))))))
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    p(2) -> 1*
    f5() -> 2*
  
  second automaton:
   final states: {3}
   transitions:
    0(15) -> 3*
    s(7) -> 8*
    s(12) -> 13*
    s(6) -> 7*
    s(9) -> 10*
    s(11) -> 12*
    s(5) -> 6*
    s(4) -> 5*
    s(8) -> 9*
    s(10) -> 11*
    s(14) -> 15*
    s(13) -> 14*
    f5() -> 4*
    15 -> 3*