YES

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

Consider the left-linear TRS R:

  E(+(x,y)) -> *(E(x),E(y))
  E(0()) -> 1()
  +(x,0()) -> x
  +(0(),x) -> x
  *(x,1()) -> x
  *(1(),x) -> x
  +(+(x,y),z) -> +(x,+(y,z))
  +(x,+(y,z)) -> +(+(x,y),z)
  +(x,y) -> +(y,x)
  *(*(x,y),z) -> *(x,*(y,z))
  *(x,*(y,z)) -> *(*(x,y),z)
  *(x,y) -> *(y,x)

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

  phi(E(+(x,y)) -> *(E(x),E(y))) = 9
  phi(E(0()) -> 1()) = 1
  phi(+(x,0()) -> x) = 3
  phi(+(0(),x) -> x) = 4
  phi(*(x,1()) -> x) = 2
  phi(*(1(),x) -> x) = 1
  phi(+(+(x,y),z) -> +(x,+(y,z))) = 13
  phi(+(x,+(y,z)) -> +(+(x,y),z)) = 15
  phi(+(x,y) -> +(y,x)) = 14
  phi(*(*(x,y),z) -> *(x,*(y,z))) = 5
  phi(*(x,*(y,z)) -> *(*(x,y),z)) = 7
  phi(*(x,y) -> *(y,x)) = 7

  psi(E(+(x,y)) -> *(E(x),E(y))) = 8
  psi(E(0()) -> 1()) = 1
  psi(+(x,0()) -> x) = 6
  psi(+(0(),x) -> x) = 12
  psi(*(x,1()) -> x) = 5
  psi(*(1(),x) -> x) = 8
  psi(+(+(x,y),z) -> +(x,+(y,z))) = 10
  psi(+(x,+(y,z)) -> +(+(x,y),z)) = 12
  psi(+(x,y) -> +(y,x)) = 11
  psi(*(*(x,y),z) -> *(x,*(y,z))) = 6
  psi(*(x,*(y,z)) -> *(*(x,y),z)) = 4
  psi(*(x,y) -> *(y,x)) = 3