YES

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

Consider the left-linear TRS R:

  a1() -> b1()
  a1() -> c1()
  b1() -> b2()
  c1() -> c2()
  a2() -> b2()
  a2() -> c2()
  b2() -> b3()
  c2() -> c3()
  a3() -> b3()
  a3() -> c3()
  b3() -> b4()
  c3() -> c4()
  a4() -> b4()
  a4() -> c4()
  b4() -> b5()
  c4() -> c5()
  a5() -> b6()
  b5() -> b6()
  c5() -> b6()

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

  phi(a1() -> b1()) = 4
  phi(a1() -> c1()) = 2
  phi(b1() -> b2()) = 1
  phi(c1() -> c2()) = 1
  phi(a2() -> b2()) = 5
  phi(a2() -> c2()) = 2
  phi(b2() -> b3()) = 1
  phi(c2() -> c3()) = 1
  phi(a3() -> b3()) = 5
  phi(a3() -> c3()) = 2
  phi(b3() -> b4()) = 1
  phi(c3() -> c4()) = 4
  phi(a4() -> b4()) = 4
  phi(a4() -> c4()) = 2
  phi(b4() -> b5()) = 1
  phi(c4() -> c5()) = 1
  phi(a5() -> b6()) = 1
  phi(b5() -> b6()) = 1
  phi(c5() -> b6()) = 1

  psi(a1() -> b1()) = 2
  psi(a1() -> c1()) = 2
  psi(b1() -> b2()) = 3
  psi(c1() -> c2()) = 1
  psi(a2() -> b2()) = 2
  psi(a2() -> c2()) = 5
  psi(b2() -> b3()) = 3
  psi(c2() -> c3()) = 1
  psi(a3() -> b3()) = 2
  psi(a3() -> c3()) = 5
  psi(b3() -> b4()) = 3
  psi(c3() -> c4()) = 1
  psi(a4() -> b4()) = 2
  psi(a4() -> c4()) = 4
  psi(b4() -> b5()) = 3
  psi(c4() -> c5()) = 1
  psi(a5() -> b6()) = 1
  psi(b5() -> b6()) = 3
  psi(c5() -> b6()) = 1