YES

1 decompositions
 #1 -----------
   1: q(q(x1,x2),x3) -> q(x1,q(x2,x3))
   2: q(x1,q(x2,x3)) -> q(q(x1,x2),x3)
   3: q(e(),x1) -> x1
   4: q(x1,e()) -> x1
   5: plus(x1,plus(x2,x3)) -> plus(plus(x1,x2),x3)
   6: plus(x1,x2) -> plus(x2,x1)
   7: plus(0(),x1) -> x1
   8: plus(s(x1),x2) -> s(plus(x1,x2))
   9: len(e()) -> 0()
  10: len(a()) -> s(0())
  11: len(q(a(),x2)) -> plus(len(a()),len(x2))

@Commutation Lemma

@Rule Labeling
--- R
   1: q(q(x1,x2),x3) -> q(x1,q(x2,x3))
   2: q(x1,q(x2,x3)) -> q(q(x1,x2),x3)
   3: q(e(),x1) -> x1
   4: q(x1,e()) -> x1
   5: plus(x1,plus(x2,x3)) -> plus(plus(x1,x2),x3)
   6: plus(x1,x2) -> plus(x2,x1)
   7: plus(0(),x1) -> x1
   8: plus(s(x1),x2) -> s(plus(x1,x2))
   9: len(e()) -> 0()
  10: len(a()) -> s(0())
  11: len(q(a(),x2)) -> plus(len(a()),len(x2))

--- S
   2: q(x1,q(x2,x3)) -> q(q(x1,x2),x3)

@Commutation Lemma

@Rule Labeling
--- R
   2: q(x1,q(x2,x3)) -> q(q(x1,x2),x3)

--- S
   1: q(q(x1,x2),x3) -> q(x1,q(x2,x3))
   3: q(e(),x1) -> x1
   4: q(x1,e()) -> x1
   5: plus(x1,plus(x2,x3)) -> plus(plus(x1,x2),x3)
   6: plus(x1,x2) -> plus(x2,x1)
   7: plus(0(),x1) -> x1
   8: plus(s(x1),x2) -> s(plus(x1,x2))
   9: len(e()) -> 0()
  10: len(a()) -> s(0())
  11: len(q(a(),x2)) -> plus(len(a()),len(x2))

@Jouannaud and Kirchner's criterion
--- R
   1: q(q(x1,x2),x3) -> q(x1,q(x2,x3))
   3: q(e(),x1) -> x1
   4: q(x1,e()) -> x1
   5: plus(x1,plus(x2,x3)) -> plus(plus(x1,x2),x3)
   6: plus(x1,x2) -> plus(x2,x1)
   7: plus(0(),x1) -> x1
   8: plus(s(x1),x2) -> s(plus(x1,x2))
   9: len(e()) -> 0()
  10: len(a()) -> s(0())
  11: len(q(a(),x2)) -> plus(len(a()),len(x2))

--- S
   1: q(q(x1,x2),x3) -> q(x1,q(x2,x3))
   3: q(e(),x1) -> x1
   4: q(x1,e()) -> x1
   5: plus(x1,plus(x2,x3)) -> plus(plus(x1,x2),x3)
   6: plus(x1,x2) -> plus(x2,x1)
   7: plus(0(),x1) -> x1
   8: plus(s(x1),x2) -> s(plus(x1,x2))
   9: len(e()) -> 0()
  10: len(a()) -> s(0())
  11: len(q(a(),x2)) -> plus(len(a()),len(x2))