YES

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

Consider the left-linear TRS R:

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

Let C be the following subset of R:

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

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.

# 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))
  dbl(x) -> +(x,x)
  +(+(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))) = 3
  phi(+(0(),y) -> y) = 2
  phi(+(s(x),y) -> s(+(x,y))) = 3
  phi(dbl(x) -> +(x,x)) = 1
  phi(+(+(x,y),z) -> +(x,+(y,z))) = 5
  phi(+(x,y) -> +(y,x)) = 5

  psi(+(x,0()) -> x) = 2
  psi(+(x,s(y)) -> s(+(x,y))) = 6
  psi(+(0(),y) -> y) = 3
  psi(+(s(x),y) -> s(+(x,y))) = 2
  psi(dbl(x) -> +(x,x)) = 1
  psi(+(+(x,y),z) -> +(x,+(y,z))) = 4
  psi(+(x,y) -> +(y,x)) = 1