YES

# Parallel rule labeling (Zankl et al. 2015).

Consider the left-linear TRS R:

  +(x,0()) -> x
  +(x,s(y)) -> s(+(x,y))
  +(0(),y) -> y
  +(s(x),y) -> s(+(x,y))
  *(x,0()) -> 0()
  *(x,s(y)) -> +(*(x,y),x)
  *(0(),y) -> 0()
  *(s(x),y) -> +(*(x,y),y)
  +(+(x,y),z) -> +(x,+(y,z))
  +(x,y) -> +(y,x)

All parallel critical peaks (except C's) are decreasing wrt rule labeling:

  phi(+(x,0()) -> x) = 2
  phi(+(x,s(y)) -> s(+(x,y))) = 4
  phi(+(0(),y) -> y) = 1
  phi(+(s(x),y) -> s(+(x,y))) = 4
  phi(*(x,0()) -> 0()) = 3
  phi(*(x,s(y)) -> +(*(x,y),x)) = 9
  phi(*(0(),y) -> 0()) = 3
  phi(*(s(x),y) -> +(*(x,y),y)) = 9
  phi(+(+(x,y),z) -> +(x,+(y,z))) = 7
  phi(+(x,y) -> +(y,x)) = 7

  psi(+(x,0()) -> x) = 4
  psi(+(x,s(y)) -> s(+(x,y))) = 8
  psi(+(0(),y) -> y) = 5
  psi(+(s(x),y) -> s(+(x,y))) = 3
  psi(*(x,0()) -> 0()) = 6
  psi(*(x,s(y)) -> +(*(x,y),x)) = 10
  psi(*(0(),y) -> 0()) = 6
  psi(*(s(x),y) -> +(*(x,y),y)) = 10
  psi(+(+(x,y),z) -> +(x,+(y,z))) = 6
  psi(+(x,y) -> +(y,x)) = 3