YES

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

Consider the left-linear TRS R:

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

Let C be the following subset of R:

  (empty)

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

  phi(+(0(),y) -> y) = 2
  phi(+(x,0()) -> x) = 1
  phi(+(s(x),y) -> s(+(y,x))) = 7
  phi(+(x,s(y)) -> s(+(x,y))) = 7
  phi(+(x,+(y,z)) -> +(+(x,y),z)) = 11
  phi(+(x,y) -> +(y,x)) = 9

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


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:

  (empty)

Let C be the following subset of R:

  (empty)

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

  (empty)

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.