YES

# Compositional parallel rule labeling (Shintani and Hirokawa 2022).

Consider the left-linear TRS R:

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

Let C be the following subset of R:

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

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

  phi(+(x,0()) -> x) = 0
  phi(+(x,s(y)) -> s(+(x,y))) = 0
  phi(+(x,+(y,z)) -> +(+(x,y),z)) = 0
  phi(+(+(x,y),z) -> +(x,+(y,z))) = 1

  psi(+(x,0()) -> x) = 0
  psi(+(x,s(y)) -> s(+(x,y))) = 0
  psi(+(x,+(y,z)) -> +(+(x,y),z)) = 0
  psi(+(+(x,y),z) -> +(x,+(y,z))) = 1


Therefore, the confluence of R follows from that of C.

# Compositional parallel critical pair system (Shintani and Hirokawa 2022).

Consider the left-linear TRS R:

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

Let C be the following subset of R:

  (empty)

The parallel critical pair system PCPS(R,C) is:

  +(y1,+(y2,0())) -> +(y1,y2)
  +(y1,+(y2,0())) -> +(+(y1,y2),0())
  +(y1,+(y2,s(x1_2))) -> +(y1,s(+(y2,x1_2)))
  +(y1,+(y2,s(x1_2))) -> +(+(y1,y2),s(x1_2))
  +(y1,+(y2,+(x1_2,x1_3))) -> +(y1,+(+(y2,x1_2),x1_3))
  +(y1,+(y2,+(x1_2,x1_3))) -> +(+(y1,y2),+(x1_2,x1_3))

All pairs in PCP(R) are joinable and PCPS(R,C)/R is terminating.
Therefore, the confluence of R follows from that of C.

# emptiness

The empty TRS is confluent.